User:Egm4313.s12.team3/Report2

=Report 2=

Problem Statement
A)Find the non-homogeneous ODE in standard form B)Find the solution in terms of the initial conditions C)Graph the solution D)Find three non-standard and non-homogeneous ODE's that are generated from the given roots Given:
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$$\displaystyle \lambda_1=-2 $$
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$$\displaystyle \lambda_2=5 $$
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$$\displaystyle y(0)=1 $$
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$$\displaystyle y'(0)=0 $$   |}
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Problem Solution
Characteristic Equation:
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$$\displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) $$
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$$\displaystyle (\lambda+2)(\lambda-5) $$
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$$\displaystyle \lambda^2-5\lambda+2\lambda-10=0 $$
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$$\displaystyle \lambda^2-3\lambda-10=0 $$ Non-Homogenous Second Order Ordinary Differential Equation:
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$$\displaystyle y''-3y'-10y=r(x) $$
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Homogenous Equation:


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$$\displaystyle y_h(x)=C_1e^{-2x}+C_2e^{5x} $$
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$$\displaystyle y'_h(x)=-2C_1e^{-2x}+5C_2e^{5x} $$ Plugging in original function from Problem Statement y(0)=1, y'(0)=0:
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$$\displaystyle y_h(0)=C_1+C_2=1 $$
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$$\displaystyle y'_h(x)=-2C_1+5C_2=0 $$
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Using Substitution To Solve:
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$$\displaystyle C_1+C_2=1 $$
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$$\displaystyle C_1=1-C_2 $$ Plugging in:
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$$\displaystyle -2C_1+5C_2=0 $$
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$$\displaystyle -2(1-C_1)+5C_2=0 $$
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$$\displaystyle -2+2C_2+5C_2=0 $$
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$$\displaystyle 2=7C_2 $$
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Value of C2:
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$$\displaystyle C_2=2/7 $$ Finding C1:
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$$\displaystyle C_1+C_2=1 $$
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$$\displaystyle C_1+2/7=1 $$
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$$\displaystyle C_1=5/7 $$
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Plugging in:


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$$ \displaystyle y(x)=\frac{5}{7} e^{-2x}+\frac{2}{7} e^{5x} $$
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Plotted Equation:
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EQUATION 1:
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$$\displaystyle 10(\lambda+2)(\lambda-5) $$
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$$\displaystyle 10\lambda^2-30\lambda-100=r(x) $$ EQUATION 2:
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$$\displaystyle 20(\lambda+2)(\lambda-5)) $$     |}
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$$\displaystyle 20\lambda^2-60\lambda-200=r(x) $$ EQUATION 3:
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$$\displaystyle 30(\lambda+2)(\lambda-5) $$
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$$\displaystyle 30\lambda^2-90\lambda-300=r(x) $$
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Solved by: Egm4313.s12.team3.guzy 3:19, 7 February 2012 (UTC)

Problem Statement
Find the solution to the L2 ODE with no excitation.
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$$\displaystyle {y}''-10{y}'+25y=r(x) $$     (2.1) Given initial conditions:
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$$\displaystyle y(0)=1 $$
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$$\displaystyle {y}'(0)=0 $$ Since there is no excitation:
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$$\displaystyle r(x)=0 $$
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Problem Solution
The Characteristic Equation
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$$\displaystyle \lambda^2+a\lambda+b=r(x)=0 $$     (2.2) From the original equation 2.1:
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$$\displaystyle a=-10 $$
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$$\displaystyle b=25 $$ Which yields:
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$$\displaystyle \lambda^2-10\lambda+25=0 $$    (2.3) Equation 2.3 can be simplified to:
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$$\displaystyle (\lambda-5)^2 $$ It has a double root;
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$$\displaystyle \lambda=5 $$ Also can be shown by discriminant:
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$$\displaystyle (-10)^2-(4)(25)=0 $$ The General Solution
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$$\displaystyle y=(c_1+c_2x)e^{5x} $$     (2.4) Take the Derivative in order to apply initial conditions
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$$\displaystyle {y}'=c_2e^{5x}+5(c_1+c_2x)e^{5x} $$    (2.5) Apply initial conditions into equations 2.4 and 2.5:
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$$\displaystyle y(0)=1=c_1 $$    (2.6)
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$$\displaystyle {y}'(0)=0=c_2+5c_1 $$    (2.7) Solving equations 2.6 and 2.7 yields:
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$$\displaystyle c_1=1 $$    (2.8)
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$$\displaystyle c_2=-5 $$    (2.9) Plugging the these values back into the general solution 2.4 yields:
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$$\displaystyle y=(1-5x)e^{5x} $$
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Plotting the obtained solution:

Solved by: Egm4313.s12.team3.landers 15:31, 6 February 2012 (UTC)

Problem Statement
Find the general solution for the given ODE and check the answer using substitution.

Part A
Given:
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$$\displaystyle y''+6y'+8.96y=0 $$    (3.1) Using the general form,
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$$\displaystyle (\lambda^2+a\lambda+b)e^{\lambda x}=0 $$    (3.2) But $$e^{\lambda x}$$ cannot equal zero. Therefore,
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$$\displaystyle \lambda^2+6\lambda+8.96=0 $$    (3.3) Using the quadratic equation, solve for $$\lambda_1$$ and $$\lambda_2$$.
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$$\displaystyle \lambda_{1,2} = -\frac{a}{2} (\frac{+}{-}) (\frac{1}{2})\sqrt{a^2-4b} = -\frac{6}{2}  (\frac{+}{-}) (\frac{1}{2})\sqrt{6^2-4(8.96} = -3 (\frac{+}{-}) 0.2 $$     (3.4) Therefore the roots of (3.3) are,
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$$\displaystyle \lambda_{1} = -2.8    \lambda_{2} = -3.2 $$    (3.5) Guessing that a solution to (3.1) will have the form,
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$$\displaystyle y=e^{(\lambda)(x)} $$    (3.6) We combine (3.5) and (3.6) to get solutions,
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$$\displaystyle y_1=e^{-2.8x} y_2=e^{-3.2x} $$    (3.7) To check if these are true solutions for (3.1) we substitute them back into the equation. First we must obtain the derivatives.
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$$\displaystyle y_1^'=-2.8e^{-2.8x} y_2^'=-3.2e^{-3.2x} y_1^{}=7.89e^{-2.8x}  y_2^{}=10.24e^{-3.2x} $$    (3.8) Substituting in the values of $$\displaystyle y_1$$ into (3.1),
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$$\displaystyle 7.89e^{-2.8x}+6(-2.8e^{-2.8x})+8.96e^{-2.8x}=16.8e^{-2.8x}-16.8e^{-2.8x}=0 $$    (3.9) Solution $$\displaystyle y_1$$ is valid. Repeating this process for $$\displaystyle y_2$$.
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$$\displaystyle 10.24e^{-3.2x}+6(-3.2e^{-3.2x})+8.96e^{-3.2x}=19.2e^{-3.2x}-19.2e^{-3.2x}=0 $$    (3.10) Solution $$\displaystyle y_2$$ is valid. Since (3.1) is a linear ODE, using superposition is valid. Therefore the solution to the ODE is,
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$$\displaystyle y(x)=c_1e^{-2.8x}+c_2e^{-3.2x} $$    (3.11)
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Part B
Given:
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$$  \displaystyle y''+4y'+(\pi^2 +4)y=0 $$ (3.12) Using the general form
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$
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Equation (3.12) becomes
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$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)=0 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{16-4(\pi^2+4)})}2 \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{16-4(\pi^2+4)})}2 $$ $$   \displaystyle \lambda_1=\frac{-4+(\sqrt{-4\pi^2)}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-(\sqrt{-4\pi^2)}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+2\pi i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=\frac{-4-2\pi i}{2} $$ $$   \displaystyle \lambda_1=-2+\pi i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-\pi i $$
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$$  \displaystyle y_1=e^{-2x} \cos{\pi x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; y_2=e^{-2x} \sin{\pi x} $$ Calculating the derivatives:
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$$  \displaystyle y=Ae^{-2x} \cos{\pi x}+Be^{-2x} \sin{\pi x} $$ $$ \displaystyle y'=[-2Ae^{-2x} \cos{\pi x}-A\pi e^{-2x} \sin{\pi x}]+[\pi Be^{-2x} \cos{\pi x}-2Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle y''=[4Ae^{-2x} \cos{\pi x}\;+2\;\pi Ae^{-2x} \sin{\pi x}\;+\;2\pi Ae^{-2x} \sin\pi x\;-\;\pi^2 Ae^{-2x} \cos{\pi x}]\;+\;[-2B\pi e^{-2x} \cos{\pi x}\;-\;\pi^2 Be^{-2x} \sin{\pi x}\;+\;4Be^{-2x} \sin{\pi x}\;-\;2\pi Be^{-2x} \cos{\pi x}] $$ Simplifying some of the terms:
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$$  \displaystyle 4y'=[-8Ae^{-2x} \cos{\pi x}-4A\pi e^{-2x} \sin{\pi x}]+[4\pi Be^{-2x} \cos{\pi x}-8Be^{-2x} \sin{\pi x}] $$ $$ \displaystyle (\pi^2 +4)y=(\pi^2 +4)Ae^{-2x} \cos{\pi x}+(\pi^2 +4)Be^{-2x} \sin{\pi x} $$
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Finally subbing the terms into the original equation:
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$$  \displaystyle y''\;+\;4y'\;+\;(\pi^2\;+\;4)y=[e^{-2x} \cos{\pi x}]\cdot[4A\;-\;\pi^2 A\;-\;2\pi B\;-\;2\pi B\;-\;8A\;+\;4\pi B\;+\;\pi^2 A\;+\;4A]\;+\;[e^{-2x} \sin{\pi x}]\cdot[2\pi A\;+\;2\pi A\;-\;\pi^2 B\;+\;4B\;-\;4\pi A\;-\;8B\;+\;\pi^2 B\;+\;4B] $$ $$ \displaystyle [e^{-2x} \cos{\pi x}]\cdot[0]+[e^{-2x} \sin{\pi x}]\cdot[0]=0 $$
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Solved by: Egm4313.s12.team3.chaffee 07:25, 8 February 2012 (UTC)

Problem Statement
For the following two problems find the general solution for the equations and then check the answer by substitution.
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$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ $$\displaystyle 10y''-32y'+25.6y=0 $$
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Part A
This problem is very similar to [| EGM4313 Team 3's Report 1 Problem 5, Part B]
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$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ (4.1) From the characteristic equation (found on K 2011 p.54 (3)):
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.2) Using (4.2) and substituting the corresponding values for a and b you get:
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$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ (4.3) Next we will check for the possibility of a Real Double Root (K 2011 p.55 "Case II") using:
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$$  \displaystyle a^2-4b=0 $$ (4.4)
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$$  \displaystyle (2\pi)^2-4(\pi^2)=0 $$ $$  \displaystyle 4\pi^2 - 4\pi^2=0 $$ $$  \displaystyle 0=0 $$ Since the Equation is a Real Double Root you now use:
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$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.5) Solving for $$\displaystyle -a/2 $$ equals:
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$$  \displaystyle \frac{-a}{2}=\frac{-2\pi}{2}=-\pi $$ (4.6) Substituting (4.6) into (4.5) leaves:
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$$  \displaystyle y=(C_1+C_2x)e^{-\pi x} $$ (4.7) Taking the first and second derivative of (4.7) gives:
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$$  \displaystyle y'=-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x} $$ (4.8)
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$$  \displaystyle y''=\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x} $$ (4.9) Finally inserting (4.7),(4.8), and (4.9) into (4.1) allows a check of the answer:
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$$  \displaystyle (\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x})+2\pi (-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x})+\pi^2 ((C_1+C_2x)e^{-\pi x})=0 $$ (4.10)
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Through simple canceling of like terms from (4.10) gives:
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$$  \displaystyle 0=0 $$
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Part B
Going over many of the same steps in Part A we have the following equation:
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$$\displaystyle 10y''-32y'+25.6=0 $$ (4.11) Which when divided by 10 equals:
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$$\displaystyle y''-3.2y'+2.56=0 $$ (4.12) From the characteristic equation (again found on K 2011 p.54 (3)):
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$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.13) Using (4.12) and substituting the corresponding values for a and b you get:
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$$  \displaystyle \lambda^2-3.2 \lambda+2.56=0 $$ (4.14) We will again check for the possibility of a Real Double Root (also on K 2011 p.55 "Case II") using:
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$$  \displaystyle a^2-4b=0 $$ (4.15)
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$$  \displaystyle (-3.2)^2-4(2.56)=0 $$ $$  \displaystyle 10.24 - 10.24=0 $$ $$  \displaystyle 0=0 $$ Since this equation is also a Real Double Root you use:
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$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.16) Solving for $$\displaystyle -a/2 $$ equals:
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$$  \displaystyle \frac{-a}{2}=\frac{-(-3.2)}{2}=1.6 $$ (4.17) Substituting (4.17) into (4.16) gives the equation:
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$$  \displaystyle y=(C_1+C_2x)e^{1.6x} $$ (4.18) Taking the first and second derivative of (4.18) gives:
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$$  \displaystyle y'=((1.6)(C_1+C_2x)+C_2)e^{1.6x} $$ (4.19)
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$$  \displaystyle y''=2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})) $$ (4.20) BY inserting (4.18),(4.19), and (4.20) into (4.11) the answer can be verified:
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$$  \displaystyle 10(2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})))-32(((1.6)(C_1+C_2x)+C_2)e^{1.6x})+25.6((C_1+C_2x)e^{1.6x})=0 $$ (4.21)
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As in Part A simple canceling of similar terms gives:
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$$  \displaystyle 0=0 $$ Solved by: Egm4313.s12.team3.droll 05:16, 8 February 2012 (UTC)
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Problem Statement
Find an ODE in the form $$y''+ay'+by=0$$ for the given basis. Part A:
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$$\displaystyle e^{2.6x}, \; e^{-4.6x} $$ Part B:
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$$\displaystyle e^{-\sqrt{5}x},\; xe^{-\sqrt{5}x} $$
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Part A
Given the general solutions of $$y_{h,n}$$


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$$\displaystyle y_{h,1}=e^{\lambda_1 x}    \;\;\;\;\;\text{and}\;\;\;\;\;        y_{h,2}=e^{\lambda_2 x} $$ We see that...
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$$\displaystyle \lambda_1 = 2.6   \;\;\;\;\;\text{and}\;\;\;\;\;        \lambda_2 = -4.3 $$ Subbing the equations above into the characteristic equation...
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$$\displaystyle (\lambda + 2.6)(\lambda - 4.3)=0 $$ Expanding that equation...
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$$\displaystyle \lambda^2 -1.7\lambda - 11.18 = 0 $$ Therefore, the ODE is...
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$$\displaystyle y'' - 1.7y' - 11.18y =0 $$
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Part B
Given the general solutions of $$y_{h,n}$$
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$$\displaystyle y_{h,1}=e^{\lambda_1 x}    \;\;\;\;\;\text{and}\;\;\;\;\;        y_{h,2}=xe^{\lambda_2 x} $$ We see that, in this case (See "Case 2" in notes - [[media:iea.s12.sec5.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 5]])...
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$$\displaystyle \lambda_1 = \lambda_2 = -\sqrt{5} $$ Subbing the equations above into the characteristic equation...
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$$\displaystyle (\lambda -\sqrt{5})(\lambda - \sqrt{5})=0 $$ Expanding that equation...
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$$\displaystyle \lambda^2 -2\sqrt{5}\lambda +5 = 0 $$ Therefore, the ODE is...
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$$\displaystyle y'' - 2\sqrt{5}y' +5y =0 $$
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Solved by: Egm4313.s12.team3.rountree 19:30, 05 February 2012 (UTC)

Problem Statement
For a spring-dashpot-mass system in series find the unknown parameters k,c, and m using the characteristic equation See Lecture 5.

Problem Solution
Figure of a spring-dashpot-mass system in series Using the final equation describing the spring-dashpot-mass system:
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$$\displaystyle f(t)=m({y_k}''+\frac{k}{c}{y_k}')+ky_k \;\;\; becomes $$ $$\displaystyle f(t)=m{y_k}''+m\frac{k}{c}{y_k}'+ky_k $$ Now using the homogeneous equation:
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$$\displaystyle (\lambda-(-3)^2) = {\lambda}^2-6\lambda+9=0 $$ So replacing $$\lambda$$ with y':
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$$\displaystyle f(t)={y''}^2+6y'+9=0 $$ So now
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$$  \displaystyle m=1 $$ $$ \displaystyle m\frac{k}{c}=6 \;\;\; \rightarrow \;\;\; \frac{(1)(9)}{c}=6 \;\;\; \rightarrow  \;\;\; 9=6c \;\;\; \rightarrow \;\;\; c=1.5 $$ $$ \displaystyle k=9 $$
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Solved by: Egm4313.s12.team3.katz 15:31, 7 February 2012 (UTC)

Problem Statement
Develop the MacLaurin series for $$e^t$$ (Part A), cos(t) (Part B), and sin(t) (Part C). See Lecture 6.

Part A
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = e^t \,\,\,\,\,\,\,, \,\,\,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f'(t) = e^t \,\,\,\,\,, \,\,\,\,\, f'(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,\,, \,\,\,\, f(0) = e^0 = 1 $$ $$\displaystyle f(t) = e^t \,\,\,, \,\,\, f(0) = e^0 = 1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
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$$\displaystyle P_n(t)= f(0)+\frac {f'(0)}{1!} (t-0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \cdots $$ $$\displaystyle P_n(t)= 1+t+\frac {1}{2} (t)^2+ \frac{1}{6} (t)^3+ \cdots $$ In Sigma notation:
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$$\displaystyle e^t=\sum_{n=0}^{\infty }\frac{t^n}{n!} $$
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Part B
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\,\, f(0) = cos(0) = 1 $$ $$\displaystyle f'(t) = -sin(t) \,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\, f(0) = sin(0) = 0 $$ $$\displaystyle f^{(4)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(5)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(5)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(6)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -cos(0) = -1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$ For this problem the general form becomes:
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$$\displaystyle P_n(t)= f(0)+ \frac{f''(0)}{2!} (t-0)^2+\frac{f^{(4)}(0)}{4!}(t-0)^4+\frac{f^{(6)}(0)}{6!}(t-0)^6\cdots $$ $$\displaystyle P_n(t)= 1-\frac{1}{2}(t)^2+ \frac{1}{24} (t)^4- \frac{1}{720} (t)^6 +\cdots $$ In Sigma notation:
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$$\displaystyle cos(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} $$
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Part C
You must find the first few non-zero derivative values to develop the Taylor series at t=0 (MacLaurin series):
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$$\displaystyle f(t) = sin(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\, f'''(0) = sin(0) = 0 $$ $$\displaystyle f'(t) = cos(t) \,\,\,\,\,\,\,\,\,\,\,\,\,, \,\,\,\,\,\,\,\, f'(0) = cos(0) = 1 $$ $$\displaystyle f(t) = -sin(t) \,\,\,\,\,\,\,, \,\,\,\,\,\,\, f(0) = -sin(0) = 0 $$ $$\displaystyle f(t) = -cos(t) \,\,\,\,\,\,, \,\,\,\,\,\, f(0) = -cos(0) = -1 $$ $$\displaystyle f^{(4)}(t) = sin(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(4)}(0) = sin(0) = 0 $$ $$\displaystyle f^{(5)}(t) = cos(t) \,\,\,\,\,\,\,\,\,, \,\,\,\, f^{(5)}(0) = cos(0) = 1 $$ $$\displaystyle f^{(6)}(t) = -sin(t) \,\,\,\,, \,\,\,\, f^{(6)}(0) = -sin(0) = 0 $$ $$\displaystyle f^{(7)}(t) = -cos(t) \,\,\,\,, \,\,\,\, f^{(7)}(0) = -cos(0) = -1 $$
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The general form of Taylor series is:
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$$\displaystyle P_n(t)= f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots $$
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For this problem the general form becomes:
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$$\displaystyle P_n(t)= \frac {f'(0)}{1!} (t-0)+\frac{f^{(3)}(0)}{3!}(t-0)^3+ \frac{f^{(5)}(0)}{5!}(t-0)^5+ \frac{f^{(7)}(0)}{7!}(t-0)^7 +\cdots $$ $$\displaystyle P_n(t)= t-\frac{1}{6}(t)^3+ \frac{1}{120} (t)^5- \frac{1}{5040} (t)^7 +\cdots $$ In Sigma notation:
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$$\displaystyle sin(t)=\sum_{n=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} $$
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Problem Statement
Find the general solution for each of the ODE's given below. Use substitution to verify the answer.


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$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2)
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Part A

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$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
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This is a homogeneous, linear, second order ODE with constant coefficients in the standard form of:


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$$\displaystyle y''+ay'+by=0 $$
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The corresponding characteristic equation would be:


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$$\displaystyle \lambda^2+\lambda+3.25=0 $$ We can see that the value of the discriminant of the characteristic equation $$\ a^2-4b $$ will be negative. This means that the roots will be complex conjugates.
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$$\displaystyle \lambda_{1,2}=-\frac{1}{2}a \pm i\omega $$ Where
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$$\displaystyle \omega^2=b-\frac{1}{4}a^2 $$ Therefore the roots of the characteristic equation are:
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$$\displaystyle \lambda_{1,2}=-\frac{1}{2} \pm i \sqrt{3} $$ Refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution (with no excitation) given below for the case of complex conjugate roots:
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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
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$$\displaystyle y=e^{-x/2}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$      (1) Below, we verify that this is indeed a solution to the original ODE.
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$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
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Before verifying that (1) satisfies the original ODE (8.1), we must calculate the first derivative:


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$$\displaystyle y'=e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$
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$$\displaystyle y'=e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B] $$ And the second derivative:
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$$\displaystyle y''=e^{-\frac{x}{2}}[-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]+\frac{1}{4}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$
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$$\displaystyle y''=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B] $$ With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B]+ e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B]+3.25e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\sqrt{3}B+\frac{1}{4}A+\sqrt{3}B-\frac{1}{2}A+3.25A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\sqrt{3}A+\frac{1}{4}B-\sqrt{3}A-\frac{1}{2}B+3.25B]
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$$
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A+\frac{1}{4}A-\frac{1}{2}A+3.25A-\sqrt{3}B+\sqrt{3}B]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[\sqrt{3}A-\sqrt{3}A+\frac{1}{4}B-\frac{1}{2}B+3.25B-3B] $$
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[0]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[0] $$
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$$\displaystyle y''+y'+3.25y=0 $$
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Since the final equation matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
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Part B

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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2) This is a homogeneous, linear, second order ODE with constant coefficients in the standard form of:
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$$\displaystyle y''+ay'+by=0 $$ The corresponding characteristic equation would be:
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$$\displaystyle \lambda^2+0.54\lambda+(0.0729+\pi)=0 $$ We can see that the value of the discriminant of the characteristic equation $$\ a^2-4b $$ will be negative. This means that the roots will be complex conjugates.
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$$\displaystyle \lambda_{1,2}=-\frac{1}{2}a \pm i\omega $$ Where
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$$\displaystyle \omega^2=b-\frac{1}{4}a^2 $$ Therefore the roots of the characteristic equation are:
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$$\displaystyle \lambda_{1,2}=-0.27 \pm i \sqrt{\pi} $$ Refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution (with no excitation) given below for the case of complex conjugate roots:
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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
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$$\displaystyle y=e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$      (2)
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Below, we verify that this is indeed a solution to the original ODE.


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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$     (8.2) Before verifying that (2) satisfies the original ODE (8.2), we must calculate the first derivative:
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$$\displaystyle y'=e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$
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$$\displaystyle y'=e^{-0.27x}cos(\sqrt{\pi}x)[\sqrt{\pi}B-0.27A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\sqrt{\pi}A-0.27B] $$ And the second derivative:
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$$\displaystyle y''=e^{-0.27x}[-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]+(0.27)^{2}e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$
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$$\displaystyle y''=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B] $$ With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A+0.54\sqrt{\pi}B-(0.54)(0.27)A+(0.0729+\pi)]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B-0.54\sqrt{\pi}A-(0.54)(0.27)B+(0.0729+\pi)B] $$
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-(0.54)(0.27)A+(0.0729+\pi)A+(0.27)^{2}A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+0.54\sqrt{\pi}B]+e^{-0.27x}sin(\sqrt{\pi}x)[0.27\sqrt{\pi}A+0.27\sqrt{\pi}A-0.54\sqrt{\pi}A-\pi B+(0.27)^{2}B-(0.54)(0.27)B+(0.0729+\pi)B] $$
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0A+0B]+e^{-0.27x}sin(\sqrt{\pi}x)[0A+0B] $$
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0]+e^{-0.27x}sin(\sqrt{\pi}x)[0] $$
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$
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Since the final equation matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
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Problem Statement
In this problem we will consider several mathematical representations of physical systems using ODE's. Given the characteristic equation (or its roots), find the general solution for the corresponding ODE. Then, use the initial conditions to determine the particular solution. Plot the solution for each.

Part A


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$$\displaystyle \lambda^{2}+4\lambda+13=0 $$     (9.1)
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Part B


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$$\displaystyle \lambda_{1,2}=-3 $$     (9.2)
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Part C


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$$\displaystyle \lambda_1=-2 \;\;,\;\; \lambda_2=5 $$     (9.3)
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Consider all of the above systems given the following initial conditions.

Part A

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$$\displaystyle \lambda^{2}+4\lambda+13=0 $$     (9.1)
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This characteristic equation would correspond to the following homogeneous ODE.


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$$\displaystyle {y}''+4{y}'+13y=0 $$
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To find the general solution to this ODE, we need to determine the roots of the characteristic equation. This is done below.


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$$  \displaystyle \lambda^2+4\lambda+13=0 $$  $$   \displaystyle \lambda_1=\frac{-4+\sqrt{16-4*13}}2 \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\;\;\;\; \lambda_2=\frac{-4-\sqrt{16-4*13}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+\sqrt{-36}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\; \lambda_2=\frac{-4-\sqrt{-36}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+6i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \lambda_2=\frac{-4-6i}{2} $$ $$   \displaystyle \lambda_1=-2+3i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-3i $$
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In this case, we have two imaginary roots. We can refer to [[media:iea.s12.sec6.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 6]] for the derivation of the form of the general solution below.


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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$
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Given the specific characteristic equation, we can apply the determined roots to get the general solution:


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$$\displaystyle y_h(x)=e^{-2x}[Acos(3x)+Bsin(3x)] $$
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Applying the initial condition $$\ y(0)=1 $$ we get:


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$$\displaystyle 1=e^{0}[Acos(0)+Bsin(0)] $$
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$$\displaystyle 1=A(1)+B(0) $$
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$$\displaystyle A=1 $$
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We now take the first derivative of the general solution:


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$$\displaystyle {y_h}'(x)=e^{-2x}[-3Asin(3x)+3Bcos(3x)]-2e^{-2x}[Acos(3x)+Bsin(3x)] $$
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Applying the initial condition $$\ y'(0)=0 $$:


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$$\displaystyle 0=e^{0}[-3Asin(0)+3Bcos(0)]-2e^{0}[Acos(0)+Bsin(0)] $$
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$$\displaystyle 0=[-3A(0)+3B(1)]-2[A(1)+B(0)] $$
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$$\displaystyle 0=3B-2A $$
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$$\displaystyle 0=3B-2(1) $$
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$$\displaystyle B=\frac{2}{3} $$
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We have now fully determined the particular solution for the given initial conditions. This is shown below.


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$$\displaystyle y_h(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)] $$
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The figure below shows a plot of the solution.

Part B

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$$\displaystyle \lambda_{1,2}=-3 $$     (9.2)
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In this instance, the roots of the characteristic equation are already provided. Although we technically do not need the corresponding characteristic equation and homogeneous ODE to proceed, they are shown below for the sake of reference.

Characteristic Equation:


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$$\displaystyle (\lambda-(-3))^2=0 $$
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$$\displaystyle (\lambda+3)^2=0 $$
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$$\displaystyle \lambda^{2}+6\lambda+9=0 $$
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Corresponding Homogeneous ODE:


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$$\displaystyle y''+6y'+9y=0 $$
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In this case, we have a double real root. We can refer to [[media:iea.s12.sec5.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 5]] for the derivation of the form of the general solution below.


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$$\displaystyle y_h(x)=c_1e^{\lambda x}+c_2xe^{\lambda x} $$
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Applying the root value to the form above, we obtain the general solution:


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$$\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x} $$
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Applying the initial condition $$\ y(0)=1 $$ we get:


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$$\displaystyle 1=c_1e^{0}+c_2(0)e^{0} $$
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$$\displaystyle 1=c_1(1)+0 $$
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$$\displaystyle c_1=1 $$
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We now take the first derivative of the general solution:


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$$\displaystyle {y_h}'(x)=-3c_1e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x} $$
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Applying the initial condition $$\ y'(0)=0 $$:


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$$\displaystyle 0=-3c_1e^{0}+c_2e^{0}-3c_2(0)e^{0} $$
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$$\displaystyle 0=-3c_1+c_2 $$
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$$\displaystyle 0=-3(1)+c_2 $$
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$$\displaystyle c_2=3 $$
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We have now fully determined the particular solution for the given initial conditions. This is shown below.


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$$\displaystyle y_h(x)=e^{-3x}+3xe^{-3x} $$
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Part C

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$$\displaystyle \lambda_1=-2 \;\;,\;\; \lambda_2=5 $$     (9.3)
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In this instance, the roots of the characteristic equation are already provided. The corresponding characteristic equation and homogeneous ODE were determined in problem 1. They are shown below.

Characteristic Equation:


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$$\displaystyle \lambda^{2}-3\lambda+-10=0 $$
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Corresponding Homogeneous ODE:


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$$\displaystyle y''-3y'-10y=0 $$
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In this case, we have two distinct, real roots. We can refer to [[media:iea.s12.sec3.djvu| Loc Vu-Quoc class EGM 4313 notes Sec 3]] for the form of the general solution below.


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$$\displaystyle y_h(x)=c_1e^{\lambda _1 x}+c_2e^{\lambda_2 x} $$
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Applying the distinct root values to the form above, we obtain the general solution:


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$$\displaystyle y_h(x)=c_1e^{-2 x}+c_2e^{5 x} $$
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Applying the initial condition $$\ y(0)=1 $$ we get:


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$$\displaystyle 1=c_1e^{0}+c_2e^{0} $$
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$$\displaystyle 1=c_1(1)+c_2 $$
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$$\displaystyle c_1+c_2=1 $$
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We now take the first derivative of the general solution:


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$$\displaystyle {y_h}'(x)=-2c_1e^{-2x}+5c_2e^{5x} $$
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Applying the initial condition $$\ y'(0)=0 $$:


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$$\displaystyle 0=-2c_1e^{0}+5c_2e^{0} $$
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$$\displaystyle 0=-2c_1+5c_2 $$
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We have 2 equations and 2 unknowns. Solving for $$\ c_1 $$:
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$$\displaystyle 0=-2c_1+5(1-c_1) $$
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$$\displaystyle 0=-7c_1+5 $$
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$$\displaystyle c_1=\frac{5}{7} $$
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Using this value to determine $$\ c_2 $$:


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$$\displaystyle 1=c_1+c_2 $$
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$$\displaystyle 1=\frac{5}{7}+c_2 $$
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$$\displaystyle c_2=\frac{2}{7} $$
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We have now fully determined the particular solution for the given initial conditions. This is shown below.


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$$\displaystyle y_h(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$
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Plot of All Solutions
Below the solutions to Parts A,B,and C have been plotted. For reference, Part A (shown in green) was the case of imaginary roots. Part B (shown in red) was the case of a double real root. Part C (shown in blue) was the case of distinct, real roots.