User:Egm4313.s12.team17/Report 2

= Problem 1 - Non-Homogeneous L2-ODE-CC =

Given
Roots of the characteristic equation:
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$$  \displaystyle \lambda_1 = -2, \lambda_2 = +5 $$     (Eq.1.1)
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Initial Conditions:
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$$  \displaystyle y(0) = 1, y'(0) = 0 $$     (Eq.1.2)
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Part 1 Problem Statement
Find the standard form and solution in terms of the initial conditions with a general excitation r(x)  for the non-homogeneous L2-ODE-CC.

Part 1 Solution
Given the roots, one can find the characteristic equation with the given general form:
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$$  \displaystyle (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$     (Eq.1.1.1)
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Substituting the corresponding values of and  into Eq. 1.1.1:


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$$  \displaystyle (\lambda - (-2))(\lambda - 5) = \lambda^2 - 3\lambda - 10 = 0 $$     (Eq.1.1.2)
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From Eq. 1.1.2, the non-homogeneous L2-ODE-CC is:
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$$  \displaystyle y'' - 3y' - 10y = r(x) $$     (Eq.1.1.3)
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To solve for the constants c_1 \, and c_2 \,, a homogeneous solution will be generated and substituted into the overall solution:
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$$  \displaystyle y(x) = y_h(x) + y_p(x) $$     (Eq.1.1.4)
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Homogeneous Solution:
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$$  \displaystyle y_h(x) = c_1e^{-2x} + c_2e^{5x} $$     (Eq.1.1.5)
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Substituting the Eq. 1.1.5 into Eq. 1.1.4:
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$$  \displaystyle y(x) = c_1e^{-2x} + c_2e^{5x} + y_p(x) $$     (Eq.1.1.6)
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Satisfying the initial conditions:
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$$  \displaystyle y(0) = 1 = c_1 + c_2 + y_p(0) $$     (Eq.1.1.7)
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$$  \displaystyle y'(0) = 0 = -2c_1 + 5c_2 + y_p'(0) $$     (Eq.1.1.8)
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Multiplying Eq. 1.1.7 by 2, adding 1.1.7 and 1.1.8, and then solve for c_2 \,:
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$$  \displaystyle c_2 = \frac{2}{7} - \frac{1}{7}[2y_p(0) + y_p'(0)] $$     (Eq.1.1.9)
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Multiplying Eq. 1.1.7 by 5, subtracting 1.1.8 and 1.1.7, and then solve for c_1 \,:
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$$  \displaystyle c_1 = \frac{5}{7} - \frac{1}{7}[5y_p(0) - y_p'(0)] $$     (Eq.1.1.10)
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Because the excitation r(x)  is equal to 0, c_1 \, and c_2 \, can be solved for:
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$$  \displaystyle c_1 = \frac{5}{7} $$     (Eq.1.1.11)
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$$  \displaystyle c_2 = \frac{2}{7} $$     (Eq.1.1.12)
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The final equation is then given by:
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$$  \displaystyle y(x) = \frac{5}{7}e^{-2x} + \frac{2}{7}e^{5x} $$     (Eq.1.1.13)
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With the corresponding plot of Eq. 1.1.13:



Part 2 Problem Statement
Admit the two values from p.3-7 (3a) as the two roots of the corresponding characteristic equation and generate three non-standard (and non-homogeneous) L2-ODE-CC.

Part 2 Solution
Three non-standard and non-homogeneous ( r(x) \ne 0 ) L2-ODE-CC:
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$$  \displaystyle 4(\lambda - (-2))(\lambda - 5) = 4\lambda^2 -28\lambda - 40 $$     (Eq.1.2.1)
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$$  \displaystyle 5(\lambda - (-2))(\lambda - 5) = 5\lambda^2 -35\lambda - 50 $$     (Eq.1.2.2)
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$$  \displaystyle 6(\lambda - (-2))(\lambda - 5) = 6\lambda^2 -42\lambda - 60 $$     (Eq.1.2.3)
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Author & Proofreaders
Author: Egm4313.s12.team17.ying 17:11, 6 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.wheeler.tw 18:00, 6 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 04:55, 8 February 2012 (UTC)

= Problem 2 - Solution for L2-ODE-CC =

Given

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$$  \displaystyle y'' - 10y' + 25y = r(x) $$     (Eq.2.1)
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Initial Conditions:
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$$  \displaystyle y(0) = 1, y'(0) = 0 $$     (Eq.2.2)
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No Excitation:
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$$  \displaystyle r(x) = 0 $$     (Eq.2.3)
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Problem Statement
For the given L2-ODE-CC (Eq. 2.1), find and plot the solution.

Solution
From section 5 of the lecture notes, pg.5.6 given the homogeneous solution as:
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$$  \displaystyle y_h(x) = c_1e^{5x} + c_2xe^{5x} $$     (Eq.2.4)
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Therefore, the solution is:
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$$  \displaystyle y(x) = c_1e^{5x} + c_2xe^{5x} + y_p(x) $$     (Eq.2.5)
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Satisfying the initial conditions:
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$$  \displaystyle y(0) = 1 = c_1 + y_p(0) $$     (Eq.2.6)
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$$  \displaystyle y'(0) = 0 = 5c_1 + \cancel{c_2*5xe^{5x}} + c_2 + y_p'(0) $$     (Eq.2.7)
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Multiplying Eq. 2.6 by 5, subtracting 2.7 from 2.6, and then solve for c_2 \,:
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$$  \displaystyle c_2 = 5 + 5 y_p(0) - y_p'(0) $$     (Eq.2.8)
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Solve for c_1 \, from Eq. 2.6:
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$$  \displaystyle c_1 = 1 - y_p(0) $$     (Eq.2.9)
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Because the excitation r(x)  is equal to 0, c_1 \, and c_2 \, can be solved for:
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$$  \displaystyle c_1 = 1 $$     (Eq.2.10)
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$$  \displaystyle c_2 = 5 $$     (Eq.2.11)
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The final equation is then:
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$$  \displaystyle y(x) = e^{5x} + 5xe^{5x} $$     (Eq.2.12)
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The plot of the solution is as follows:



Author & Proofreaders
Author: Egm4313.s12.team17.ying 00:19, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.hintz 04:25, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 05:19, 8 February 2012 (UTC)

= Problem 3 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

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$$  \displaystyle y''+6y'+8.96y=0 $$     (Eq.3.1.1)
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Part 1 Problem Statement
Find a general solution for Eq. 3.1.1 and confirm the general solution using substitution.

Part 1 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 3.1.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)


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$$  \displaystyle \lambda^2+6\lambda+8.96\lambda=0 $$     (Eq.3.1.2)
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Since Eq. 3.1.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.3.1.3)
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When we apply the coefficients of Eq. 3.1.2 (where a=1, b=6 , and c=8.96 ) to Eq. 3.1.3 we get:
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$$  \displaystyle \lambda=\frac{-6\pm \sqrt{6^2-4(1)(8.96)}}{2(1)} $$     (Eq.3.1.4)
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Which can be simplified to yield the root(s) of the characteristic equation as shown below:
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$$  \displaystyle \lambda=\frac{-6\pm \sqrt{36-35.84}}{2} $$     (Eq.3.1.5)
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$$  \displaystyle \lambda=\frac{-6\pm \sqrt{0.16}}{2} $$     (Eq.3.1.6)
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$$  \displaystyle \lambda=\frac{-6\pm 0.4}{2} $$     (Eq.3.1.7)
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$$  \displaystyle \lambda=-3\pm 0.2 $$     (Eq.3.1.8)
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At this point we can see that there are two roots to the characteristic equation due to the presence of the \pm sign. The two roots will be called \lambda_1 and \lambda_2 and are shown below:
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$$  \displaystyle \lambda_1=-3+0.2 $$     (Eq.3.1.9)
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$$  \displaystyle \lambda_2=-3-0.2 $$     (Eq.3.1.10)
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The simplified roots are shown below:
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$$  \displaystyle \lambda_1=-2.8 $$     (Eq.3.1.11)
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$$  \displaystyle \lambda_2=-3.2 $$     (Eq.3.1.12)
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Eq. 3.1.11 and Eq. 3.1.12 indicate that this equation has two distinct, real roots and as such, the general solution will be in the following form:
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$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$     (Eq.3.1.13)
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Where c_1 and c_2 are arbitrary constants and \lambda_1 and \lambda_2 are defined in Eq. 3.1.11 and Eq. 3.1.12, respectively.

The general solution is shown below:
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$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$     (Eq.3.1.14)
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Now we're going to check through substitution that the general solution we've obtained in Eq. 3.1.14 is valid:
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$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$     (Eq.3.1.14)
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$$  \displaystyle y'=-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x} $$     (Eq.3.1.15)
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$$  \displaystyle y''=7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} $$     (Eq.3.1.16)
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If substituting Eq. 3.1.14, Eq. 3.1.15, and Eq. 3.1.16 into Eq. 3.1.1 and does not violate Eq. 3.1.1, then we can accept Eq. 3.1.14 to be the general solution.
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$$  \displaystyle 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} + 6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+8.96(c_1e^{-2.8x}+c_2e^{-3.2x})=0 $$     (Eq.3.1.17)
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We will simplify this equation until we can conclusively show whether or not the general solution is valid:
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$$  \displaystyle 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} - 16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x}=0 $$     (Eq.3.1.18)
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$$  \displaystyle -8.96c_1e^{-2.8x}-8.96c_2e^{-3.2x} +8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x}=0 $$     (Eq.3.1.19)
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$$  \displaystyle 0=0 $$     (Eq.3.1.20)
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Now we have proven that Eq. 3.1.14 is indeed the general solution to Eq. 3.1.1.

Part 2 Given

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$$  \displaystyle y''+4y'+( \pi^2 + 4 )y=0 $$     (Eq.3.2.1)
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Part 2 Problem Statement
Find a general solution for Eq. 3.2.1 and confirm the general solution using substitution.

Part 2 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 3.2.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)
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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)\lambda=0 $$     (Eq.3.2.2)
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Since Eq. 3.2.2 is of the form we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.3.2.3)
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When we apply the coefficients of Eq. 3.2.2 (where a=1, b=4 , and c=\pi^2+4 ) to Eq. 3.2.3 we get:
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$$  \displaystyle \lambda=\frac{-4\pm \sqrt{4^2-4(1)(\pi^2+4)}}{2(1)} $$     (Eq.3.2.4)
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Which can be simplified to yield the root(s) of the characteristic equation as shown below:
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$$  \displaystyle \lambda=\frac{-4\pm \sqrt{16-16-4\pi^2}}{2} $$     (Eq.3.2.5)
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$$  \displaystyle \lambda=\frac{-4\pm \sqrt{-4\pi^2}}{2} $$     (Eq.3.2.6)
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$$  \displaystyle \lambda=\frac{-4\pm 2 \pi i}{2} $$     (Eq.3.2.7)
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$$  \displaystyle \lambda=-2\pm \pi i $$ (Eq.3.2.8)
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At this point we can see that there are two roots to the characteristic equation. The two roots will be called \lambda_1 and \lambda_2 and are shown below:
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$$  \displaystyle \lambda_1=-2+ \pi i $$ (Eq.3.2.9)
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$$  \displaystyle \lambda_2=-2- \pi i $$ (Eq.3.2.10)
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Eq. 3.2.9 and Eq. 3.2.10 indicate that this equation has two complex conjugate roots and as such, the general solution will be in the following form:
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$$  \displaystyle y=e^{ax}(c_1\cos wx+c_2\sin wx) $$     (Eq.3.2.11)
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Where c_1 and c_2 are arbitrary constants and a is the real component of the complex conjugate and w is the imaginary component of the complex conjugate roots are defined in Eq. 3.2.9 and Eq. 3.2.10, respectively.

The general solution is shown below:
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$$  \displaystyle y=e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ (Eq.3.2.12)
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Now we're going to check through substitution that the general solution we've obtained in Eq. 3.2.12 is valid:
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$$  \displaystyle y=e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ (Eq.3.2.12)
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$$  \displaystyle y'=-2e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)+e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x) $$ (Eq.3.2.13)
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$$  \displaystyle y''=4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-2e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)-2e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x) $$ $$  \displaystyle +e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x) $$ (Eq.3.2.14)
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Eq. 3.2.14 can be further simplified as shown below:
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$$  \displaystyle y''=4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-4e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)+e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x) $$ (Eq.3.2.15)
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If substituting Eq. 3.2.12, Eq.3.2.13, and Eq. 3.2.15 into Eq. 3.2.1 and does not violate Eq. 3.2.1, then we can accept Eq. 3.1.12 to be the general solution.
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$$  \displaystyle 4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-4e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)+e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x)+4(-2e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ $$  \displaystyle    +e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x))+( \pi^2 + 4 )(e^{-2x}(c_1\cos \pi x+c_2\sin \pi x))=0 $$     (Eq.3.1.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle (-4e^{-2x}+4e^{-2x})(c_1\cos \pi x+c_2\sin \pi x)=0 $$     (Eq.3.2.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.3.2.18)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now we have proven that Eq. 3.2.12 is indeed the general solution to Eq. 3.2.1.

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 15:12, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 16:05, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 06:21, 8 February 2012 (UTC)

= Problem 4 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

 * {| style="width:100%" border="0"

$$  \displaystyle y''+2 \pi y'+\pi^2 y=0 $$     (Eq.4.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Problem Statement
Find a general solution for Eq. 4.1.1 and confirm the general solution using substitution.

Part 1 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 4.1.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+2 \pi\lambda+\pi^2\lambda=0 $$     (Eq.4.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since Eq. 4.1.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.4.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

When we apply the coefficients of Eq. 4.1.2 (where a=1, b=2\pi , and c=\pi^2 ) to Eq. 4.1.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(2 \pi )\pm \sqrt{(2\pi)^2-4(1)(\pi^2)}}{2(1)} $$     (Eq.4.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(2 \pi )\pm \sqrt{4\pi^2-4\pi^2}}{2} $$     (Eq.4.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-2 \pi\pm \sqrt{0}}{2} $$     (Eq.4.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=-\pi $$     (Eq.4.1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From Eq. 4.1.7 we can observe that there is a real double root to the characteristic equation. As such the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{\lambda x} $$ (Eq.4.1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Where c_1 and c_2 are arbitrary constants and \lambda is defined in Eq. 4.1.7, above.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 4.1.9 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y'=c_2e^{-\pi x}-\pi(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y''=-c_2\pi e^{-\pi x}-c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

If substituting Eq. 4.1.9, Eq. 4.1.10, and Eq. 4.1.11 into Eq. 4.1.1 and does not violate Eq. 4.1.1, then we can accept Eq. 4.1.9 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle -c_2\pi e^{-\pi x}-c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x}+2 \pi (c_2e^{-\pi x}-\pi(c_1+c_2 x)e^{-\pi x})+\pi^2 ((c_1+c_2 x)e^{-\pi x})=0 $$     (Eq.4.1.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle -2c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x}+2 \pi c_2e^{-\pi x}-2 \pi^2(c_1+c_2 x)e^{-\pi x}+\pi^2 ((c_1+c_2 x)e^{-\pi x})=0 $$     (Eq.4.1.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \pi^2(c_1+c_2 x)e^{-\pi x}-\pi^2(c_1+c_2 x)e^{-\pi x}=0 $$     (Eq.4.1.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.4.1.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now we have proven that Eq. 4.1.9 is indeed the general solution to Eq. 4.1.1.

Part 2 Given

 * {| style="width:100%" border="0"

$$  \displaystyle 10y''-32y'+25.6y=0 $$     (Eq.4.2.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 2 Problem Statement
Find a general solution for Eq. 4.2.1 and confirm the general solution using substitution.

Part 2 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 4.2.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)


 * {| style="width:100%" border="0"

$$  \displaystyle 10\lambda^2-32\lambda+25.6\lambda=0 $$     (Eq.4.2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since Eq. 4.2.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.4.2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

When we apply the coefficients of Eq. 4.2.2 (where a=10, b=-32 , and c=25.6 ) to Eq. 4.2.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{32\pm \sqrt{(-32)^2-4(10)(25.6)}}{2(10)} $$     (Eq.4.2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{32\pm \sqrt{1024-1024}}{20} $$     (Eq.4.2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{32\pm \sqrt{0}}{20} $$     (Eq.4.2.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=1.6 $$     (Eq.4.2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From Eq. 4.2.7 we can observe that there is a real double root to the characteristic equation. As such the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{\lambda x} $$ (Eq.4.2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Where c_1 and c_2 are arbitrary constants and \lambda is defined in Eq. 4.2.7, above.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 4.2.9 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y'=c_2e^{1.6 x}+1.6(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y''=1.6c_2 e^{1.6 x}+1.6 c_2 e^{1.6 x}+1.6^2(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

If substituting Eq. 4.2.9, Eq. 4.2.10, and Eq. 4.2.11 into Eq. 4.2.1 and does not violate Eq. 4.2.1, then we can accept Eq. 4.2.9 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle 10(1.6c_2 e^{1.6 x}+1.6 c_2 e^{1.6 x}+1.6^2(c_1+c_2 x)e^{1.6 x})-32(c_2e^{1.6 x}+1.6(c_1+c_2 x)e^{1.6 x})+25.6((c_1+c_2 x)e^{1.6 x})=0 $$     (Eq.4.2.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle 16c_2 e^{1.6 x}+16 c_2 e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}-32c_2e^{1.6 x}-51.2(c_1+c_2 x)e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}=0 $$     (Eq.4.2.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle 16c_2 e^{1.6 x}-16 c_2 e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}-25.6(c_1+c_2 x)e^{1.6 x}=0 $$     (Eq.4.2.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.4.2.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now we have proven that Eq. 4.2.9 is indeed the general solution to Eq. 4.2.1.

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 15:26, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.hintz 17:31, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 08:09, 8 February 2012 (UTC)

= Problem 5 - Reverse Engineering = For problem reference.

Part 1 Given

 * {| style="width:100%" border="0"

$$  \displaystyle e^{2.6x}, e^{-4.3x} $$     (Eq.5.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Problem Statement
Find an ODE for the given basis when.

Part 1 Solution
The roots of Eq. 5.1.1 are:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1 = 2.6 $$     (Eq.5.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_2 = -4.3 $$     (Eq.5.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To obtain the formula of the factored quadratic, we set the roots (Eq. 5.1.2 and 5.1.3) equal to zero:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda_1 - 2.6) = 0 $$     (Eq.5.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda_2 + 4.3) = 0 $$     (Eq.5.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To obtain the original function, we multiply Eq. 5.1.4 and 5.1.5 together to yield:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda_1 - 2.6)(\lambda_2 + 4.3) = 0 $$     (Eq.5.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 + (4.3\lambda) -(2.6\lambda) - 11.18 =0 $$     (Eq.5.1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Simplifying Eq. 5.1.7:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 + 1.7\lambda - 11.18=0 $$     (Eq.5.1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Converting the characteristic equation to an ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime} + 1.7y^{\prime} - 11.18y=0 $$     (Eq.5.1.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part 2 Given

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-\sqrt{5}x}, x e^{-\sqrt{5}x} $$     (Eq.5.2.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 2 Problem Statement
Find an ODE for the given basis when.

Part 2 Solution
The root of Eq. 5.2.1 is:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda = -\sqrt{5} $$     (Eq.5.2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To obtain the formula of the factored quadratic, we set Eq. 5.2.2 equal to zero:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda + \sqrt{5}= 0 $$     (Eq.5.2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To obtain the original function, we square Eq. 5.2.3 to yield the answer:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 +2\sqrt{5}\lambda + 5 =0 $$     (Eq.5.2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Converting the characteristic equation to an ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime} +2\sqrt{5}y^\prime + 5y =0 $$     (Eq.5.2.5)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Author & Proofreaders
Author: Egm4313.s12.team17.hintz 10:00, 3 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 19:32, 3 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 08:31, 8 February 2012 (UTC)

= Problem 6 - Spring-Dashpot-Mass System in Series =

Given



 * k \, is the spring constant
 * c \, is the damping constant
 * y(t) is the displacement of the mass

The equation for this system is:
 * {| style="width:100%" border="0"

$$  \displaystyle my_k^{\prime\prime}+\frac{mk}{c}y_k^\prime+ky_k=f(t) $$     (Eq.6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Parameters to this equation can be found by the characteristic equation for a double root if given \lambda.

Problem Statement
Find the values for the coefficients m, k, and c. Using \lambda=-3.

Solution
When \lambda=-3, then the characteristic equation becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda + 3)^2=\lambda^2+6\lambda+9=0 $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Comparing Eq 6.1 and Eq 6.2 we can set the coefficients of each equation equal to each other:
 * {| style="width:100%" border="0"

$$  \displaystyle m = 1 $$     (Eq.6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \frac{mk}{c} = 6 $$     (Eq.6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle k = 9 $$     (Eq.6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We can see that coefficients m and k have already been solved.

Using Eq 6.4, c can be solved as well:
 * {| style="width:100%" border="0"

$$  \displaystyle c =\frac{mk}{6} $$     (Eq.6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle c = \frac{(1)(9)}{6} $$     (Eq.6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle c = \frac{3}{2} $$     (Eq.6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we have:
 * {| style="width:100%" border="0"

$$  \displaystyle m=1 $$     (Eq.6.9)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle k=9 $$     (Eq.6.10)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle c=\frac{3}{2} $$     (Eq.6.11)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Now, the equation for this system is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_k^{\prime\prime}+6y_k^\prime+9y_k=f(t) $$     (Eq.6.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Author & Proofreaders
Author: Egm4313.s12.team17.Li Kelvin Li 06:12, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.axelrod.a 15:57, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 10:56, 8 February 2012 (UTC)

= Problem 7 - MacLaurin Series =

Given
By definition a Taylor series about x=a  is defined as:
 * {| style="width:100%" border="0" align="left"

$$  \displaystyle f(t)= \sum_{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(t-a)^n $$       (Eq.7.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">
 * }

Where f ^{(n)} represents the order of the derivative.

For a MacLaurin Series a=0. The formula for a MacLaurin Series becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)= \sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}(t)^n $$     (Eq.7.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for e^t.

Part 1 Solution

 * {| style="width:100%" border="0"

$$  \displaystyle f(t)= e^t $$     (Eq.7.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=0 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(0)}(0)=e^{(0)}=1 $$     (Eq.7.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}e^t=e^0=1 $$     (Eq.7.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}e^t=e^0=1 $$     (Eq.7.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}e^t=e^0=1 $$     (Eq.7.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}e^t=e^0=1 $$     (Eq.7.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}e^t=e^0=1 $$     (Eq.7.1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Apply Eq. 7.1.2 - 7.1.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=\frac{1}{0!}t^0+\frac{1}{1!}t^1+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4+\frac{1}{5!}t^5 $$     (Eq.7.1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the Maclaurin Series expansion for f(t)=e^t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120} .... $$ (Eq.7.1.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Determined from the pattern of Eq. 7.1.8, the summation for a Taylor Series approximation of f(t)=e^t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=e^t=\sum_{n=0}^{\infty }\frac{t^n}{n!} $$     (Eq.7.1.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part 2 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for \cos t.

Part 2 Solution

 * {| style="width:100%" border="0"

$$  \displaystyle f(t)= cos(t) $$     (Eq.7.2.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=0 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(0)}(0)=cos(0)=1 $$     (Eq.7.2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}cos(0)=-sin(0)=0 $$     (Eq.7.2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}cos(0)=-cos(0)=-1 $$     (Eq.7.2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}cos(0)=sin(0)=0 $$     (Eq.7.2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}sin(0)=cos(0)=1 $$     (Eq.7.2.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}cos(0)=-sin(0)=0 $$     (Eq.7.2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Apply Eq. 7.2.2 - 7.2.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=\frac{1}{0!}t^0+\frac{0}{1!}t^1-\frac{1}{2!}t^2+\frac{0}{3!}t^3+\frac{1}{4!}t^4+\frac{0}{5!}t^5 $$     (Eq.7.2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the Maclaurin Series expansion for f(t)=\cos t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}-\frac{t^6}{720}+\frac{t^8}{40320} .... $$ (Eq.7.2.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Determined from the pattern of Eq. 7.2.8, the summation for a Taylor Series approximation of f(t)=\cos t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=cos(t)=\sum_{n=0}^{\infty }\frac{[(-1)^n]t^{2n}}{(2n)!} $$     (Eq.7.2.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part 3 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for \sin t.

Part 3 Solution

 * {| style="width:100%" border="0"

$$  \displaystyle f(t)= sin(t) $$     (Eq.7.3.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=0 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(0)}(0)=sin(0)=0 $$     (Eq.7.3.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}sin(0)=cos(0)=1 $$     (Eq.7.3.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}sin(0)=-sin(0)=0 $$     (Eq.7.3.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}sin(0)=-cos(0)=-1 $$     (Eq.7.3.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}sin(0)=sin(0)=0 $$     (Eq.7.3.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}sin(0)=cos(0)=1 $$     (Eq.7.3.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Apply Eq. 7.3.2 - 7.3.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=\frac{0}{0!}x^0+\frac{1}{1!}x^1+\frac{0}{2!}x^2-\frac{1}{3!}x^3+\frac{0}{4!}x^4+\frac{1}{5!}x^5 $$     (Eq.7.3.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the Maclaurin Series expansion for f(t)=\sin t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} .... $$     (Eq.7.3.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Determined from the pattern of Eq. 7.3.8, the summation for a Taylor Series approximation of f(t)=\sin t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=sin(t)=\sum_{n=0}^{\infty }\frac{[(-1)^n]t^{1+2n}}{(1+2n)!} $$     (Eq.7.3.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 19:22, 7 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.Li 21:12, 7 February 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.hintz 22:31, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 09:21, 8 February 2012 (UTC)

= Problem 8 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}+y^\prime+3.25y=0 $$     (Eq.8.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Problem Statement
Find an ODE for the given basis when.

Part 1 Solution
Use the following to generate the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substitute Eq. 8.1.2 into Eq. 8.1.1
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+\lambda +3.25=0 $$     (Eq.8.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To solve the characteristic equation, apply the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (Eq.8.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(1)\pm\sqrt{(1)^2-4(1)(3.25)}}{2(1)} $$     (Eq.8.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-1\pm\sqrt{-12}}{2} $$     (Eq.8.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-1\pm i \sqrt{12}}{2} $$     (Eq.8.1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_{1}=\frac{-1+ i \sqrt{12}}{2} $$     (Eq.8.1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_{2}=\frac{-1- i \sqrt{12}}{2} $$     (Eq.8.1.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.8.1.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(C_1\cos\omega x+C_2\sin\omega x) $$ (Eq.8.1.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then, the general solution will be the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y= e^{-\frac{1}{2}x}(C_1\cos\sqrt{12}x+C_2\sin\sqrt{12}x) $$     (Eq.8.1.12)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

In order to check the general solution, take the first and second derivative of Eq. 8.1.12 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime= -\frac{1}{2}e^{-\frac{1}{2} x }(C_{1}\cos\sqrt{12}x+C_{2}\sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x) $$     (Eq.8.1.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=\frac{1}{4}e^{-\frac{1}{2} x }(C_{1}\cos\sqrt{12}x+C_{2}\sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x) $$ $$ -\frac{1}{2}e^{-\frac{1}{2} x }(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x)+e^{-\frac{1}{2} x }(-12C_{1}\cos\sqrt{12}x-12C_{2}\sin\sqrt{12}x) $$     (Eq.8.1.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substitute Eq. 8.1.12, 8.1.13, and 8.1.14 into Eq. 8.1.1.

The final result of that substitution is:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.8.1.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Part 2 Given

 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}+0.54y^\prime+(0.0729+\pi)y=0 $$     (Eq.8.2.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 2 Problem Statement
Find an ODE for the given basis when.

Part 2 Solution
Use the following to generate the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substitute Eq. 8.2.2 into Eq. 8.2.1
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+0.54\lambda +(0.0729+\pi)=0 $$     (Eq.8.2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

To solve the characteristic equation, apply the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(0.54)\pm\sqrt{(0.54)^2-4(1)(0.0729+\pi)}}{2(1)} $$     (Eq.8.2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-0.54\pm\sqrt{-\pi}}{2} $$     (Eq.8.2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-0.54\pm i\sqrt{\pi}}{2} $$     (Eq.8.2.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_{1}=-0.27+ i \sqrt{\pi} $$     (Eq.8.2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_{2}=-0.27- i \sqrt{\pi} $$     (Eq.8.2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.8.2.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(C_1\cos\omega x+C_2\sin\omega x) $$ (Eq.8.2.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then, the general solution will be the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y= e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x) $$     (Eq.8.2.11)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

In order to check the general solution, take the first and second derivative of Eq. 8.2.11 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime= -0.27e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x)+e^{-0.27x}(-\sqrt{\pi}C_{1}\sin\pi x+\sqrt{\pi}C_{2}\cos \pi x) $$ (Eq.8.2.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=0.0729e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x)-0.27e^{-0.27x} (-\sqrt{\pi}C_{1}\sin\sqrt{\pi}x+\sqrt{\pi}C_{2}\cos\sqrt{\pi}x) $$
 * style="width:95%" |
 * style="width:95%" |

$$ -0.27e^{-0.27x}(-\sqrt{\pi}C_{1}\sin\pi x+\sqrt{\pi}C_{2}\cos\pi x)+e^{-0.27x}(-\pi C_{1}\cos\sqrt{\pi}x-\pi C_{2}\sin\sqrt{\pi}x) $$     (Eq.8.2.13)
 * <p style="text-align:right">
 * }

Substitute Eq. 8.2.11, 8.2.12, and 8.2.13 into Eq. 8.2.1.

The final result of that substitution is:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.8.2.14)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 19:40, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.deaver.md 10:23, 8 February 2012 (UTC) Editor: Egm4313.s12.team17.deaver.md 10:23, 8 February 2012 (UTC)

= Problem 9 - Three Cases of Damping=

Part 1 Given
Initial Conditions:
 * {| style="width:100%" border="0"

$$  \displaystyle y(0)=1,y^\prime(0)=0 $$     (Eq.9.1.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

No excitation:
 * {| style="width:100%" border="0"

$$  \displaystyle r(x)=0 $$     (Eq.9.1.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Characteristic Equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+4\lambda+13=0 $$     (Eq.9.1.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Part 1 Problem Statement
Find the solution for the L2-ODE-CC that correspond to Eq. 9.1.3 and plot the solution.

Part 1 Solution
Solve for the roots in Eq. 9.1.3 by using the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{4^2-4(1)(13)}}{2(1)} $$     (Eq.9.1.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{-36}}{2} $$     (Eq.9.1.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=-2\pm3i $$     (Eq.9.1.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.9.1.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(A\cos\omega x+B\sin\omega x) $$ (Eq.9.1.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Using the values from Eq. 9.1.6, substitute them into Eq. 9.1.8:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-2x}(A\cos3 x+B\sin3 x) $$ (Eq.9.1.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Take the derivative of Eq. 9.1.9 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime=e^{-2x}(-2A\cos3 x - 3A\sin3 x - 2B\sin3 x + 3B\cos3 x) $$ (Eq.9.1.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solve for A \, and B \, by using the initial conditions (Eq. 9.1.1):
 * {| style="width:100%" border="0"

$$  \displaystyle 1=e^{-2(0)}(A\cos3(0)+B\sin3(0)) $$     (Eq.9.1.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle A=1 $$     (Eq.9.1.12)
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$$  \displaystyle 0=e^{-2(0)}(-2A\cos3 (0) - 3A\sin3 (0) - 2B\sin3 (0) + 3B\cos3 (0)) $$     (Eq.9.1.13)
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$$  \displaystyle 0=-2A + 3B $$     (Eq.9.1.14)
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$$  \displaystyle B=\frac{2}{3} $$     (Eq.9.1.15)
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Therefore, the solution for this L2-ODE-CC that uses the initial condition shown above is the following:
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$$  \displaystyle y=e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x) $$ (Eq.9.1.16)
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In order to check if this solution is correct, we can substitute it in the L2-ODE-CC standard form.

The standard form of the L2-ODE-CC:
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$$  \displaystyle y^{\prime\prime}+4y^\prime+13y=r(x) $$     (Eq.9.1.17)
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Use Eq. 9.1.2 to simplify Eq. 9.1.17:
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$$  \displaystyle y^{\prime\prime}+4y^\prime+13y=0 $$     (Eq.9.1.18)
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Take the derivative of Eq. 9.1.16 twice with respect to x \,:
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$$  \displaystyle y^{\prime}=-\frac{13}{3}e^{-2x}\sin3x $$     (Eq.9.1.19)
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$$  \displaystyle y^{\prime\prime}=\frac{26}{3}e^{-2x}\sin3x-13e^{-2x}\cos3x $$     (Eq.9.1.20)
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Next, take Eq. 9.1.16,9.1.19, and 9.1.20 and plug it into Eq. 9.1.18:
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$$  \displaystyle [\frac{26}{3}e^{-2x}\sin3x - 13e^{-2x}\cos3x] + 4[-\frac{13}{3}e^{-2x}\sin3x] + 13[e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x)] = 0 $$     (Eq.9.1.21)
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$$  \displaystyle \frac{26}{3}e^{-2x}\sin3x - 13e^{-2x}\cos3x - \frac{52}{3}e^{-2x}\sin3x + 13e^{-2x}\cos3 x + \frac{26}{3}e^{-2x}\sin3 x = 0 $$     (Eq.9.1.22)
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From Eq. 9.1.22, we can see that everything negates one another; therefore, Eq. 9.1.16 is a correct.

The plot of Eq. 9.1.16 is shown below:



Part 2 Given

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$$  \displaystyle y = \frac{5}{7}e^{-2x} + \frac{2}{7}e^{5x} $$     (Eq.1.1.13)
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$$  \displaystyle y_k^{\prime\prime}+6y_k^\prime+9y_k=f(t) $$     (Eq.6.12)
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$$  \displaystyle y=e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x) $$ (Eq.9.1.16)
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Part 2 Problem Statement
Take the equations that are shown in the given and superpose them.

Part 2 Solution
Need to solve for the solution for Eq. 6.12 that meet the initial condition from part 1 (Eq. 9.1.1).

Therefore, convert Eq. 6.12 into the characteristic equation:
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$$  \displaystyle \lambda^2+6\lambda+9=0 $$     (Eq.9.2.1)
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Then, use the quadratic formula to get the roots:
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$$  \displaystyle \lambda=\frac{-6\pm \sqrt{6^2-4(1)(9)}}{2(1)} $$     (Eq.9.2.2)
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$$  \displaystyle \lambda=\frac{-6\pm \sqrt{0}}{2} $$     (Eq.9.2.3)
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$$  \displaystyle \lambda=-3 $$     (Eq.9.2.2)
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Since, the roots are in the following form:
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$$  \displaystyle \lambda=a $$     (Eq.9.2.3)
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Therefore, the general solution will take the form:
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$$  \displaystyle y=e^{ax}(A+B x) $$ (Eq.9.2.4)
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Using the values from Eq. 9.2.2, substitute them into Eq. 9.2.4:
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$$  \displaystyle y=e^{-3x}(A+B x) $$ (Eq.9.2.5)
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Take the derivative of Eq. 9.1.9 with respect to x \,:
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$$  \displaystyle y^\prime=-3e^{-3x}(A+B x)+Be^{-3x} $$     (Eq.9.2.6)
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Solve for A \, and B \, by using the initial conditions (Eq. 9.1.1):
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$$  \displaystyle 1=e^{-3(0)}(A+B (0)) $$     (Eq.9.2.7)
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$$  \displaystyle A=1 $$     (Eq.9.2.8)
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$$  \displaystyle 0=-3e^{-3(0)}(A+B (0))+Be^{-3(0)} $$     (Eq.9.2.9)
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$$  \displaystyle B=3 $$     (Eq.9.2.10)
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Therefore, the solution for this L2-ODE-CC that uses the initial condition shown above is the following:
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$$  \displaystyle y=e^{-3x}(1+3 x) $$ (Eq.9.2.11)
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Now, the superpose plot shown below:



Overdamping is the distinct real roots ODE. (Eq.1.1.13)

Critical damping is real double root ODE. (Eq.9.2.11)

Underdamping is the complex conjugate roots ODE. (Eq.9.1.16)

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 11:57, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.ying 15:23, 8 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 11:57, 8 February 2012 (UTC)

= References =

= Contributing Team Members =