User:Egm4313.s12.team1.armanious/team1/R2

=Report 2=

Statement
Given the two roots and the initial conditions:


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$$  \displaystyle \lambda_1=-2, \lambda_2 = +5 $$     (1.0)
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$$  \displaystyle y(0)=1, y'(0)=0 $$     (1.1)
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Part 1. Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x).\!$$

Consider no excitation and plot the solution:


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$$  \displaystyle r(x) = 0 $$     (1.2)
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Part 2. Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $$\lambda_1=-2, \lambda_2 = +5\!$$ as the two roots of the corresponding characteristic equation.

Solution
Part 1.

Characteristic Equation:
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$$  \displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) = 0 $$     (1.3)
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$$  \displaystyle (\lambda-(-2))(\lambda-5) = 0 $$     (1.4)
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$$  \displaystyle \lambda^2-3\lambda-10=0 $$     (1.5) Non-Homogeneous L2-ODE-CC:
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$$  \displaystyle y''-3y'-10y=r(x)\rightarrow standard form $$     (1.6)
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Homogenous Solution:
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$$  \displaystyle y_h(x)=c_1e^{-2x} + c_2e^{5x} $$     (1.7) Overall Solution:
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$$  \displaystyle y(x)=c_1e^{-2x} + c_2e^{5x}+y_p(x) $$     (1.8)
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$$  \displaystyle y'(x)=-2c_1e^{-2x} + 5c_2e^{5x}+y'_p(x) $$     (1.9) Satisfy Initial Conditions:
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$$  \displaystyle y(0)=1=c_1 + c_2 +y_p(0) $$     (1.10)
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$$  \displaystyle y'(0)=0=-2c_1 + 5c_2 +y'_p(0) $$     (1.11) No excitation:
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$$  \displaystyle r(x) = 0 \rightarrow y_p(x)=0 \rightarrow y'_p(x)=0 $$     (1.12) From (1.10):
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$$  \displaystyle c_1+c_2 = 1 \rightarrow c_1=1-c_2 $$     (1.13) From (1.11):
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$$  \displaystyle -2c_1+5c_2 = 0 $$     (1.14)
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$$  \displaystyle 5c_2=2c_1 $$     (1.15)
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$$  \displaystyle \frac{5}{2}c_2=c_1 $$     (1.16) Plug (1.13) into (1.16):
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$$  \displaystyle \frac{5}{2}c_2=1-c_2 $$     (1.17) Solve for $$c_2\!$$:
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$$  \displaystyle c_2=\frac{2}{7} $$     (1.18) Plug (1.18) into (1.13) and solve for $$c_1\!$$:
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$$  \displaystyle c_1=\frac{5}{7} $$     (1.19) Therefore, the final solution in terms of the initial conditions and the general excitation $$r(x)\!$$ is:
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$$  \displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$     (1.20)
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Plot of Solution:

Figure 1

Part 2.

Three non-standard and non-homogeneous solutions using the same roots given above:

1.
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$$  \displaystyle 2(\lambda-(-2))(\lambda-5) = 0 $$     (1.21)
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$$  \displaystyle 2\lambda^2-6\lambda-20=0 $$     (1.22)
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2.
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$$  \displaystyle 3(\lambda-(-2))(\lambda-5) = 0 $$     (1.23)
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$$  \displaystyle 3\lambda^2-9\lambda-30=0 $$     (1.24)
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3.
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$$  \displaystyle 4(\lambda-(-2))(\lambda-5) = 0 $$     (1.25)
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$$  \displaystyle 4\lambda^2-12\lambda-40=0 $$     (1.26)
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Author
Solved and Typed By ---Egm4313.s12.team1.wyattling 20:26, 3 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.essenwein 20:26, 3 February 2012 (UTC)

Statement
Given initial conditions $$y(0) = 1, y'(0) = 0\!$$

No excitation:

$$r(x) = 0\!$$

Find and plot the solution for L2-ODE-CC(4) p.5-5


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$$ \displaystyle y''-10y+25y=r(x) $$     (2.1)
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Solution
The characteristic equation for this L2-ODE-CC is:


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$$ \displaystyle \lambda^2-10\lambda+25=0 $$     (2.2) With roots:
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$$ \displaystyle (\lambda-5)^2 = 0 $$     (2.3) Note: this characteristic equation has a double root, so the general form for the general solution is:
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$$ \displaystyle y(t)=C_1e^{\lambda x}+C_2xe^{\lambda x} $$ (2.4) Substituting the root values from Eq. (2.3) we see the specific general solution is:
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$$ \displaystyle y(t)=C_1e^{5x}+C_2xe^{5x} $$     (2.5) Using Eq. (2.5) and our Initial Value Condition (IVC) of $$y(0)=1\!$$:
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$$ \displaystyle y(0)=1=C_1e^{5(0)}+C_2(0)e^{5(0)} $$     (2.6) Therefore, $$C_1=1\!$$:
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In order to use the IVC of $$y'(0)=0\!$$, $$y'(t)\!$$ must be determined from Eq. (2.5):


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$$ \displaystyle y'(t)=C_15e^{5x}+C_2e^{5x}(1+5x) $$     (2.7) Substituting the value of $$C_1\!$$ and using the IVC $$y'(0)=0\!$$:
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$$ \displaystyle y'(0)=0=C_15e^{5(0)}+C_2e^{5(0)}(1+5(0)) $$     (2.8) Therefore, $$C_2=-5\!$$
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Substituting $$C_1\!$$ and $$C_2\!$$ into Eq. (2.5) a solution is found:


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$$ \displaystyle y(t)=e^{5x}-5xe^{5x} $$     (2.9)
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Plot of Solution:

Figure 1

Author
Solved and Typed By - --Egm4313.s12.team1.durrance 00:04, 8 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.armanious 03:34, 8 February 2012 (UTC)

Statement
From K 2011 pg. 59 #3,4

Find the General Solution to the two ODEs below. (See Sec.5 Pg.5-6 notes)


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$$  \displaystyle y''+6y'+8.96y=0 $$
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$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$
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Solution
Given:

The Homogeneous equation is given:


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$$  \displaystyle y''+6y'+8.96y=0 $$     (3.0)
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The first step is to change the equation into the characteristic equation.


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$$  \displaystyle \lambda ^2+6\lambda +8.96=0 $$     (3.1)
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Next use either completing the square or the quadratic formula to solve for $$\lambda\!$$ (the roots of the characteristic equation). In this case completing the square is used.


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$$  \displaystyle (\lambda ^2+6\lambda +9)+8.96=0+9 $$     (3.2)
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$$  \displaystyle (\lambda +3)^2=.04 $$     (3.3)
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The roots of the characteristic equation:


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$$  \displaystyle \lambda =-3\pm .2 $$     (3.4)
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Now that the roots are known we can use the Homogeneous solutions in this equation:


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$$  \displaystyle y=C_{1}e^{\lambda x}+C_{2}e^{\lambda x} $$ (3.5)
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Finally plug in the known roots and this is the final answer and general solution:


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$$  \displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$     (3.6)
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To prove the solution is correct we can take the solution y, the first derivative, and the second derivative and substitute them into the original homogeneous equation. If the equation equals zero, the solution is correct.

The First Derivative of the solution:
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$$  \displaystyle y'=-2.8C_{1}e^{-2.8x}-3.2C_{2}e^{-3.2} $$     (3.7)
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The Second Derivative of the solution:
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$$  \displaystyle y''=7.84C_{1}e^{-2.8x}+10.24C_{2}e^{-3.2} $$     (3.8)
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Once the solutions have been substituted and the arithmetic is done the substituted solution does equal zero so the solution is correct.


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$$  \displaystyle (7.84C_{1}e^{-2.8x}+10.24C_{2}e^{-3.2})+6(-2.8C_{1}e^{-2.8x}-3.2C_{2}e^{-3.2})+8.96(C_{1}e^{-2.8x}+C_{2}e^{-3.2})=0 $$     (3.9)
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The Homogeneous equation is given:


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$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$     (3.10)
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Again, the first step is to convert to the characteristic equation.


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$$  \displaystyle \lambda ^2+4\lambda +(\pi ^2+4)=0 $$     (3.11)
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In order to solve for the roots ($$\lambda\!$$) either use the quadratic formula or completing the square. Once again in this case completing the square was used to solve for $$\lambda\!$$.


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$$  \displaystyle (\lambda ^2+4\lambda +4)+(\pi ^2+4)=0+4 $$     (3.12)
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$$  \displaystyle (\lambda +2)^2=4-(\pi ^2+4) $$     (3.13)
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$$  \displaystyle (\lambda +2)^2=-\pi ^2 $$     (3.14)
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The root takes the square root of a negative one, therefore the complex number $$i\!$$ is in the solution. The roots of the characteristic equation is shown:


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$$  \displaystyle \lambda =-2\pm \pi i $$ (3.15)
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For complex roots the Homogeneous solution is different and the root must broken into components and plugged into the complex Homogeneous solution equation below.


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$$  \displaystyle \lambda =a\pm bi $$ (3.16)
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$$  \displaystyle y=C_{1}e^{ax}cos(bx)+C_{2}e^{ax}sin(bx) $$     (3.17)
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When plugging in $$\lambda\!$$ the final answer and general solution is:

$$  \displaystyle y=C_{1}e^{-2x}cos(\pi x)+C_{2}e^{-2x}sin(\pi x) $$ (3.18)
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Once again use substitution to prove the solution is correct. First take the first and second derivative of the solution using the product rule.

The first derivative:


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$$  \displaystyle y'=-2C_{1}e^{-2x}cos(\pi x)-\pi C_{1}e^{-2x}sin(\pi x)-2C_{2}e^{-2x}sin(\pi x)+\pi C_{2}e^{-2x}cos(\pi x) $$ (3.19)
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The second derivative:


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$$  \displaystyle y''=4C_{1}e^{-2x}cos(\pi x)+2\pi C_{1}e^{-2x}sin(\pi x)+2\pi C_{1}e^{-2x}sin(\pi x)-\pi ^{2} C_{1}e^{-2x}cos(\pi x)+4C_{2}e^{-2x}sin(\pi x)-2\pi C_{2}e^{-2x}cos(\pi x)-2\pi C_{2}e^{-2x}cos(\pi x)-\pi ^{2}C_{2}e^{-2x}sin(\pi x) $$ (3.20)
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Second Derivative simplified:


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$$  \displaystyle y''=(4-\pi ^{2}) C_{1}e^{-2x}cos(\pi x)+4\pi C_{1}e^{-2x}sin(\pi x)-4\pi C_{2}e^{-2x}cos(\pi x)+(4-\pi ^{2})C_{2}e^{-2x}sin(\pi x) $$ (3.21)
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When substituted in the the original Homogeneous equation the result is zero. The solution is correct.


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$$  \displaystyle [(4-\pi ^{2}) C_{1}e^{-2x}cos(\pi x)+4\pi C_{1}e^{-2x}sin(\pi x)-4\pi C_{2}e^{-2x}cos(\pi x)+(4-\pi ^{2})C_{2}e^{-2x}sin(\pi x)]+4[-2C_{1}e^{-2x}cos(\pi x)-\pi C_{1}e^{-2x}sin(\pi x)-2C_{2}e^{-2x}sin(\pi x)+\pi C_{2}e^{-2x}cos(\pi x)]+(\pi ^2 +4)[C_{1}e^{-2x}cos(\pi x)+C_{2}e^{-2x}sin(\pi x)]=0 $$     (3.22)
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Author
Solved and typed by - Egm4313.s12.team1.stewart 02:03, 4 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri 21:22, 5 February 2012 (UTC)

Statement
K 2011 p.59 problems 5,6.

Find a general solution. Check your answer by substitution. Problem 5:
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$$  \displaystyle y''+2 \pi y'+ \pi ^2y=0 $$
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Problem 6:
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$$  \displaystyle 10y''-32y'+25.6y=0 $$
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Solution
Problem 5:

The Homogeneous equation in standard form is given:


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$$  \displaystyle y''+2 \pi y'+ \pi ^2y=0 $$     (4.1.0)
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The corresponding characteristic equation then looks like this:


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$$  \displaystyle \lambda ^2+2 \pi \lambda + \pi ^2=0 $$     (4.1.1)
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By factoring, the characteristic equation (4.1.1) simplifies to:


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$$  \displaystyle (\lambda + \pi) ^2=0 $$     (4.1.2)
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Using (4.1.2), we can see that the equation has a repeated root where $$ \displaystyle \lambda $$ equals $$ \displaystyle -\pi $$.

Based on the value of $$ \displaystyle \lambda $$, the general solution looks like this:


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$$  \displaystyle y=C_{1}e^{-\pi x}+C_{2}x e^{-\pi x} $$ (4.1.3)
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To check the general solution, we take it's derivative twice and plug it into the given homogeneous equation. The first and second derivatives of the general solution (4.1.3) are shown below:


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$$  \displaystyle y'= -\pi C_{1}e^{-\pi x} + -\pi C_{2} x e^{-\pi x} + C_{2}e^{-\pi x} $$ (4.1.4)
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$$  \displaystyle y''= \pi ^2 C_{1}e^{-\pi x} + \pi ^2 C_{2} x e^{-\pi x} + (-\pi) C_{2}e^{-\pi x} + (-\pi)C_{2}e^{-\pi x} $$ (4.1.5)
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By picking reasonable x, C1, and C2 values, we can check that there is agreement between the homogeneous solution and the general solution. For this case, we assume x=1, C1=3, and C2=2. Plugging in these values into (4.1.3),(4.1.4), and (4.1.5), the general solution, and the derivatives equal this:


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$$  \displaystyle y= 3 e^{-\pi} + 2 e^{-\pi} = 5e^{-\pi} $$     (4.1.6)
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$$  \displaystyle y'= -\pi 3 e^{-\pi} + -\pi 2 e^{-\pi} + 2 e^{-\pi}= -5 \pi e^{-\pi} +2 e^{-\pi} $$     (4.1.7)
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$$  \displaystyle y''= \pi ^2 3 e^{-\pi} + \pi ^2 2 e^{-\pi} + (-\pi) 2 e^{-\pi} + (-\pi)2 e^{-\pi}= 5 \pi ^2 e^{-\pi} - 4 \pi e^{-\pi} $$     (4.1.8)
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Plugging in equations (4.1.6), (4.1.7), and (4.1.8) into the original homogeneous (4.1.0) yield this result:


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$$  \displaystyle 5 \pi ^2 e^{-\pi} - 4 \pi e^{-\pi} +2 \pi( -5 \pi e^{-\pi} +2 e^{-\pi})+ \pi ^2(5e^{-\pi})=0 $$     (4.1.9)
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(4.1.9) is a correct statement, and thus he general solution (4.1.3) is a correct answer.

Problem 6:

The Homogeneous equation is given:


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$$  \displaystyle 10y''-32 y'+ 25.6y=0 $$     (4.2.0)
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The corresponding characteristic equation in standard form then looks like this:


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$$  \displaystyle \lambda ^2-3.2 \lambda + 2.56=0 $$     (4.2.1)
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By factoring, the characteristic equation (4.2.1) simplifies to:


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$$  \displaystyle (\lambda -1.6) ^2=0 $$     (4.2.2)
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Using (4.2.2), we can see that the equation has a repeated root where $$ \displaystyle \lambda $$ equals 1.6.

Based on the value of $$ \displaystyle \lambda $$, the general solution looks like this:


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$$  \displaystyle y=C_{1}e^{1.6 x}+C_{2}x e^{1.6x} $$     (4.2.3)
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To check the general solution, we take it's derivative twice and plug it into the given homogeneous equation. The first and second derivatives of the general solution (4.2.3) are shown below:


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$$  \displaystyle y'= 1.6 C_{1}e^{-\pi x} + 1.6 C_{2} x e^{1.6 x} + C_{2}e^{1.6 x} $$ (4.2.4)
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$$  \displaystyle y''= 1.6 ^2 C_{1}e^{1.6 x} + 1.6 ^2 C_{2} x e^{1.6 x} + 1.6 C_{2}e^{1.6 x} + 1.6 C_{2}e^{1.6 x} $$ (4.2.5)
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By picking reasonable x, C1, and C2 values, we can check that there is agreement between the homogeneous solution and the general solution. For this case, we assume x=1, C1=3, and C2=2. Plugging in these values into (4.2.3),(4.2.4), and (4.2.5), the general solution, and the derivatives equal this:


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$$  \displaystyle y= 3 e^{1.6} + 2 e^{1.6} = 5e^{1.6} $$     (4.2.6)
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$$  \displaystyle y'= (1.6) 3 e^{1.6} + (1.6) 2 e^{1.6} + 2 e^{1.6}= 5 (1.6) e^{1.6} +2 e^{1.6} $$     (4.2.7)
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$$  \displaystyle y''= (1.6) ^2 3 e^{1.6} + (1.6) ^2 2 e^{1.6} + (1.6) 2 e^{1.6} + (1.6)2 e^{1.6}= 5 (1.6) ^2 e^{1.6} + 4 (1.6) e^{1.6} $$     (4.2.8)
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Plugging in equations (4.2.6), (4.2.7), and (4.2.8) into the original homogeneous (4.2.0) yield this result:


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$$  \displaystyle 10(5 (1.6) ^2 e^{1.6} + 4 (1.6) e^{1.6}) - 32 (5 (1.6) e^{1.6} +2 e^{1.6})+ 25.6(5e^{1.6})=0 $$     (4.2.9)
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(4.2.9) is a correct statement, and thus he general solution (4.2.3) is a correct answer.

Author
Solved and Typed By - Egm4313.s12.team1.silvestri 21:23, 5 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.stewart 22:24, 5 February 2012 (UTC)

Statement
K 2011 p.59 problems 16, 17.

Find an ODE of the form $$y''+ay'+by=0\!$$ given the following solutions:

$$16)    e^{2.6x}, e^{-4.3x}\!$$

$$17)    e^{-\sqrt{5}x}, xe^{-\sqrt{5}x}\!$$

Solution
For problem 16:


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$$  \displaystyle \lambda_{1}=2.6, \lambda_{2}=-4.3 $$     (5.0)
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To find the coefficients of the ODE:


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$$  \displaystyle (\lambda-2.6)(\lambda+4.3)=\lambda^2-1.7\lambda-11.18 $$     (5.1)
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These are the coefficients of the homogeneous L2-ODE-CC corresponding to the given solution.


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$$  \displaystyle y''-1.7y'-11.18y=0 $$     (5.2)
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For problem 17:

The solution of the characteristic equation is a double root:


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$$  \displaystyle \lambda_1=\lambda_2=-\sqrt{5} $$     (5.3)
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To find the coefficients:
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$$  \displaystyle (\lambda+\sqrt{5})^2=\lambda^2+2\sqrt{5}\lambda+5 $$     (5.4)
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Thus the homogeneous L2-ODE-CC corresponding to the given solution is:
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$$  \displaystyle y''+2\sqrt{5}y'+5y=0 $$     (5.5)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious 19:57, 5 February 2012 (UTC)

Reviewed By ---Egm4313.s12.team1.durrance 19:07, 8 February 2012 (UTC)

Statement
Realize a spring-dashpot-mass system in series as shown in Fig. p.1-4 with the same characteristic as in (3) p.5-5, i.e., with double real root $$\lambda =-3$$ , i.e., find the values for the parameters $$k$$, $$c$$, $$m$$.

Solution
From p.1-4:
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$$  \displaystyle my_{k}^{''}+m\frac{k}{c}y_{k}^{'}+k{{y}_{k}}=f(t) $$     (6.0)
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$$  \displaystyle y_{k}^{''}+\frac{k}{c}y_{k}^{'}+\frac{k}{m}{{y}_{k}}=f(t) $$     (6.1)
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Double real root implies the characteristic equation
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$$  \displaystyle {{(\lambda -(-3))}^{2}}=0 $$     (6.2)
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$$  \displaystyle {{\lambda }^{2}}+6\lambda +9=0 $$     (6.3)
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The linear second order differential equation with characteristic equation (6.2) must have the form


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$$  \displaystyle y''+6y'+9y=f(t) $$     (6.4) By inspection we see that
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$$  \displaystyle m=1 $$     (6.5)
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$$  \displaystyle \frac{k}{c}=6 $$     (6.6)
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$$  \displaystyle \frac{k}{m}=9 $$     (6.7) By substitution, we find that
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$$  \displaystyle c=\frac{3}{2} $$     (6.8)
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Final Solution


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$$  \displaystyle m=1 $$ $$   \displaystyle k=9 $$ $$   \displaystyle c=\frac{3}{2} $$     (6.9)
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Author
Solved and Typed By - Egm4313.s12.team1.essenwein 16:54, 29 January 2012 (UTC) Reviewed By - --Egm4313.s12.team1.wyattling 19:15, 8 February 2012 (UTC)

Statement
Develop the MacLaurin Series (Taylor Series at $$ t = 0\! $$) for $$ e^t,\cos t,\sin t\! $$ as discussed in [[media:iea.s12.sec6.djvu| p. 6-6. ]]

Solution
The general form of a Taylor series is:


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$$  \displaystyle f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(t-a)^n $$     (7.0)
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When the series is centered around 0 ( $$a = 0\!$$ ), then the Maclaurin series is derived:


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$$  \displaystyle f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}t^n $$     (7.1)
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Maclaurin series for: $$e^t\!$$


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$$  \displaystyle f(t)=e^t $$     (7.2)
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$$f'(t)=e^t\!$$

$$f''(t)=e^t\!$$

$$f^{(3)}(t)=e^t\!$$

$$f^{(4)}(t)=e^t\!$$


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$$  \displaystyle e^t=\sum_{n=0}^{\infty}(\frac{e^0}{0!}t^0+\frac{e^0}{1!}t^1+\frac{e^0}{2!}t^2+\frac{e^0}{3!}t^3+...) $$     (7.3)
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$$  \displaystyle e^t=\sum_{n=0}^{\infty}(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+...)=\sum_{n=0}^{\infty}\frac{t^n}{n!} $$     (7.4)
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Therefore:


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$$  \displaystyle e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!} $$     (7.5)
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Maclaurin series for: $$\sin t\!$$


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$$  \displaystyle f(t)=\sin t $$ (7.6)
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$$f'(t)=\cos t\!$$

$$f''(t)=-\sin t\!$$

$$f^{(3)}(t)=-\cos t\!$$

$$f^{(4)}(t)=\sin t\!$$


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$$  \displaystyle \sin t=\sum_{n=0}^{\infty}(\frac{\sin 0}{0!}t^0+\frac{\cos 0}{1!}t^1+\frac{-\sin 0}{2!}t^2+\frac{-\cos 0}{3!}t^3+\frac{\sin 0}{4!}t^4+\frac{\cos 0}{5!}t^5+\frac{-\sin 0}{6!}t^6+\frac{-\cos 0}{7!}t^7+...) $$     (7.7)
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$$  \displaystyle \sin t=\sum_{n=0}^{\infty}(0+\frac{t^1}{1!}-0-\frac{t^3}{3!}+0+\frac{t^5}{5!}-0-\frac{t^7}{7!}+...) $$     (7.8)
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Let $$k\!$$ be defined such that:


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$$  \displaystyle n=2k+1 $$     (7.9)
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Then the above simplifies to:


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$$  \displaystyle \sin t=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}t^{2k+1} $$     (7.10)
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Maclaurin series for: $$\cos t\!$$


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$$  \displaystyle f(t)=\cos t $$ (7.11)
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$$f'(t)=-\sin t\!$$

$$f''(t)=-\cos t\!$$

$$f^{(3)}(t)=\sin t\!$$

$$f^{(4)}(t)=\cos t\!$$


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$$  \displaystyle \cos t=\sum_{n=0}^{\infty}(\frac{\cos 0}{0!}t^0+\frac{-\sin 0}{1!}t^1+\frac{-\cos 0}{2!}t^2+\frac{\sin 0}{3!}t^3+\frac{\cos 0}{4!}t^4+\frac{-\sin 0}{5!}t^5+\frac{-\cos 0}{6!}t^6+\frac{\sin 0}{7!}t^7+\frac{\cos 0}{8!}t^8+...) $$     (7.12)
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$$  \displaystyle \cos t=\sum_{n=0}^{\infty}(1-\frac{t^2}{2!}+0+\frac{t^4}{4!}-0-\frac{t^6}{6!}+0+\frac{t^8}{8!}+...) $$     (7.13)
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Let $$k\!$$ be defined such that:


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$$  \displaystyle n=2k $$     (7.14)
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Then the above simplifies to:


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$$  \displaystyle \cos t=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}t^{2k} $$     (7.15)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious 22:54, 5 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.silvestri 19:33, 8 February 2012 (UTC)

Statement
K 2011 p.60 Problems 8,15 Find a general solution to the equations, and check answers by substitution. K 2011 p.59 Problem 8
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$$  \displaystyle y''+y'+3.25y=0 $$ K 2011 p.59 Problem 15
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$$  \displaystyle y''+0.54y'+(0.0729+\pi )y=0 $$
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Solution
K 2011, p.59 problem 8
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$$  \displaystyle y''+y'+3.25y=0 $$     (8.0)
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The first step is to change the equation into the characteristic equation.


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$$  \displaystyle \lambda ^2+\lambda +3.25=0 $$     (8.1)
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Now we use the quadratic formula to solve for $$\lambda\!$$


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$$  \displaystyle \lambda =\frac {-b \pm \sqrt{b^2 - 4ac}}{2a} $$     (8.2)
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Plugging in our coefficients we get the complex conjugate roots:
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$$  \displaystyle \lambda_{1,2} = \frac {-1 \pm i\sqrt{12}}{2} $$     (8.3)
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This can be rewritten into the form:


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$$  \displaystyle \lambda_{1,2} = \frac {-1}{2}a \pm i\sqrt{3} $$     (8.4) Plugging into the following standard equation:
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$$  \displaystyle y = e^{-ax}(A\cos wx+B\sin wx) $$     (8.5) Resulting in our final answer:
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$$  \displaystyle y = e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x) $$     (8.6) In order to check our answer, we will use substitution. In order to do so, we must derive our answer twice and plug them back into our original equation. If the equation is still equal to zero, then our answer is correct. First derivative:
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$$  \displaystyle y' = -\frac{1}{2}e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)+e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x) $$     (8.7) Second derivative:
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$$  \displaystyle y'' = -\frac{1}{4}e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)-\frac{1}{2}e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x)-\frac{1}{2}e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x) $$     (8.8) Plugging into the original equation we get:
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$$  \displaystyle -\frac{1}{4}e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)-\frac{1}{2}e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x)-\frac{1}{2}e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x)+-\frac{1}{2}e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x)+e^{-\frac{x}{2}}(-\sqrt{3}A\sin \sqrt{3}x+\sqrt{3}B\cos \sqrt{3}x) $$ $$  \displaystyle +3.25(e^{-\frac{x}{2}}(A\cos \sqrt{3}x+B\sin \sqrt{3}x))=0 $$     (8.9) Because our equation is still equal to zero, our answer is correct.
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K 2011, p.59 problem 15
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$$  \displaystyle y''+.54y'+(0.0729+\pi )y=0 $$     (8.10)
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The first step is to change the equation into the characteristic equation.


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$$  \displaystyle \lambda ^2+.54\lambda +(0.0729+\pi )=0 $$     (8.11)
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Now we use the quadratic formula to solve for $$\lambda\!$$


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$$  \displaystyle \lambda =\frac {-b \pm \sqrt{b^2 - 4ac}}{2a} $$     (8.12)
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Plugging in our coefficients we get the complex conjugate roots:
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$$  \displaystyle \lambda_{1,2} = \frac {-1 \pm i\sqrt{12.566}}{2} $$     (8.13)
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This can be rewritten into the form:


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$$  \displaystyle \lambda_{1,2} = \frac {-1}{2}a \pm i\sqrt{\pi } $$     (8.14) Plugging into the following standard equation:
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 * }
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$$  \displaystyle y = e^{-ax}(A\cos wx+B\sin wx) $$     (8.15) Resulting in our final answer:
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 * }
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$$  \displaystyle y = e^{-0.27x}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x) $$     (8.16) In order to check our answer, we will use substitution. In order to do so, we must derive our answer twice and plug them back into our original equation. If the equation is still equal to zero, then our answer is correct. First derivative:
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$$  \displaystyle y' = -0.27e^{-0.27x}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x)+e^{-0.27x}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x) $$     (8.17) Second derivative:
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$$  \displaystyle y'' = -0.0729e^{-0.27}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x)-0.27e^{-0.27}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x)-0.27e^{-0.27}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x) $$     (8.18) Plugging into the original equation we get:
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$$  \displaystyle -0.0729e^{-0.27}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x)-0.27e^{-0.27}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x)-0.27e^{-0.27}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x)+0.54(-0.27e^{-0.27x}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x) $$ $$  \displaystyle +e^{-0.27x}(-\sqrt{\pi }A\sin \sqrt{\pi }x+\sqrt{\pi }B\cos \sqrt{\pi }x)+(0.0729+\pi )(e^{-0.27x}(A\cos \sqrt{\pi }x+B\sin \sqrt{\pi }x))=0 $$      (8.19) Because our equation is still equal to zero, our answer is correct.
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Author
Solved and Typed By --Egm4313.s12.team1.rosenberg 05:50, 8 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.armanious 14:51, 8 February 2012 (UTC)

Statement
As found in Sec. 6 p.6-7, find and plot the solution for the L2-ODE-CC corresponding to $$ \lambda^2 + 4\lambda + 13 = 0\!$$

Initial conditions:

$$ y(0) = 1, y'(0) = 0\!$$

No excitation:

$$ r(x) = 0\!$$

In another Fig., superpose 3 Figs.:(a) this Fig., (b) the Fig. in R2.6 Sec. 5 p.5-6, (c) the Fig. in R2.1 Sec. 3 p.3-7

Solution
The general equation, which also represents the characteristic equation is:
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$$  \displaystyle \lambda^2 + 4\lambda + 13 = 0 $$     (9.1)
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Note: The discriminant of the characteristic equation is negative which gives roots of the form:


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$$  \displaystyle \lambda=-\frac{1}{2}a\pm bi $$ (9.2)
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The general equation for this case of L2-ODE-CC is:


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$$  \displaystyle y(x)=C_1e^{ax}\cos bx+C_2e^{ax}\sin bx $$ (9.3)
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By applying the quadratic equation to the characteristic equation, we see that the roots are:


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$$  \displaystyle \lambda =-2\pm (3)i $$     (9.4)
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This yields a general solution of:

$$  \displaystyle y(x)=C_1e^{-2x}\cos (3x)+C_2e^{-2x}\sin (3x) $$     (9.5) By applying the Initial Value Conditions (IVC) $$\displaystyle y(0)=1$$ to Eq. (9.5):
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$$  \displaystyle y(0)=1=C_1e^{-2(0)}\cos (3(0))+C_2e^{-2(0)}\sin (3(0)) $$     (9.6) Therefore, $$C_1=1\!$$
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To use the IVC $$\displaystyle y'(0)=0$$, the derivative of Eq. (9.5): :{| style="width:100%" border="0" $$  \displaystyle y'(x)=e^{-2x}\cos(3x)(3C_2-2C_1)+e^{-2x}(\sin(3x)(-3C_1-2C_2) $$     (9.7) Evaluating Eq. (9.7) for IVC $$\displaystyle y'(0)=0$$ and $$\displaystyle C_1=1$$:
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$$  \displaystyle y'(0)=0=e^{-2(0)}\cos(3(0))(3C_2-2(1))+e^{-2(0)}(\sin(3(0))(-3(1)-2C_2) $$     (9.8) Therefore, $$C_2=\frac{2}{3}\!$$
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Substituting $$C_1\!$$ and $$C_2\!$$ into Eq. (9.5) yields the solution: :{| style="width:100%" border="1" $$  \displaystyle y(x)=e^{-2x}\cos (3x)+\frac {2}{3}e^{-2x}\sin (3x) $$     (9.9) Plot of Solution (a):
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Figure 1

There is no plot in R2.6 Sec. 5 p.5-6

Plot of Solution (c):

Figure 2

Author
Solved and Typed By ---Egm4313.s12.team1.durrance 01:59, 8 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.armanious 17:35, 8 February 2012 (UTC)