User:Egm4313.s12.team3.pliskow/R2work

=Problem 8: Finding the General Solution of Homogeneous ODE's=

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) 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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$$\displaystyle \lambda^2+\lambda+3.25=0 $$ 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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$$\displaystyle y=e^{-x/2}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$ 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) Before verifying that (1) satisfies the original ODE (8.1), we must calculate the first derivative:
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$$\displaystyle y'=e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (2)
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$$\displaystyle y'=e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B] $$    (3) And the second derivative:
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$$\displaystyle y''=e^{-\frac{x}{2}}[-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]+\frac{1}{4}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (4)
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$$\displaystyle y''=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B]+ e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B]+3.25e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (6)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\sqrt{3}B+\frac{1}{4}A+\sqrt{3}B-\frac{1}{2}A+3.25A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\sqrt{3}A+\frac{1}{4}B-\sqrt{3}A-\frac{1}{2}B+3.25B]
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$$    (7)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A+\frac{1}{4}A-\frac{1}{2}A+3.25A-\sqrt{3}B+\sqrt{3}B]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[\sqrt{3}A-\sqrt{3}A+\frac{1}{4}B-\frac{1}{2}B+3.25B-3B] $$    (8)
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$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[0]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[0] $$    (8)
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$$\displaystyle y''+y'+3.25y=0 $$    (8)
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Since (8) 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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$$\displaystyle y=e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$
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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 (1) satisfies the original ODE (8.2), we must calculate the first derivative:
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$$\displaystyle y'=e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$    (2)
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$$\displaystyle y'=e^{-0.27x}cos(\sqrt{\pi}x)[\sqrt{\pi}B-0.27A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\sqrt{\pi}A-0.27B] $$    (3) And the second derivative:
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$$\displaystyle y''=e^{-0.27x}[-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]-0.27e^{-0.27x}[-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)]+(0.27)^{2}e^{-0.27x}[Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)] $$    (4)
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$$\displaystyle y''=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+(0.27)^{2}A+0.54\sqrt{\pi}B-(0.54)(0.27)A+(0.0729+\pi)]+e^{-0.27x}sin(\sqrt{\pi}x)[-\pi B+0.27\sqrt{\pi}A+0.27\sqrt{\pi}A+(0.27)^{2}B-0.54\sqrt{\pi}A-(0.54)(0.27)B+(0.0729+\pi)B] $$    (6)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[-\pi A-(0.54)(0.27)A+(0.0729+\pi)A+(0.27)^{2}A-0.27\sqrt{\pi}B-0.27\sqrt{\pi}B+0.54\sqrt{\pi}B]+e^{-0.27x}sin(\sqrt{\pi}x)[0.27\sqrt{\pi}A+0.27\sqrt{\pi}A-0.54\sqrt{\pi}A-\pi B+(0.27)^{2}B-(0.54)(0.27)B+(0.0729+\pi)B] $$    (7)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0A+0B]+e^{-0.27x}sin(\sqrt{\pi}x)[0A+0B] $$    (8)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=e^{-0.27x}cos(\sqrt{\pi}x)[0]+e^{-0.27x}sin(\sqrt{\pi}x)[0] $$    (8)
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$$\displaystyle y''+0.54y'+(0.0729+\pi)y=0 $$    (8)
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Since (8) 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) Consider all of the above systems given the following initial conditions.
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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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$$\displaystyle {y}''+4{y}'+13y=0 $$ 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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$$  \displaystyle \lambda^2+4\lambda+13=0 $$  $$   \displaystyle \lambda_1=\frac{-4+\sqrt{16-4*13}}2 \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\;\;\;\; \lambda_2=\frac{-4-\sqrt{16-4*13}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+\sqrt{-36}}2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\;\;\;\; \lambda_2=\frac{-4-\sqrt{-36}}2 $$ $$   \displaystyle \lambda_1=\frac{-4+6i}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \lambda_2=\frac{-4-6i}{2} $$ $$   \displaystyle \lambda_1=-2+3i \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \lambda_2=-2-3i $$ 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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$$\displaystyle y_h(x)=e^{-ax/2}[Acos(wx)+Bsin(wx)] $$ Given the specific characteristic equation, we can apply the determined roots to get the general solution:
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$$\displaystyle y_h(x)=e^{-2x}[Acos(3x)+Bsin(3x)] $$ 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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$$\displaystyle A=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=e^{-2x}[-3Asin(3x)+3Bcos(3x)]-2e^{-2x}[Acos(3x)+Bsin(3x)] $$ 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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$$\displaystyle B=\frac{2}{3} $$ We have now fully determined the particular solution for the given initial conditions. This is shown below.
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$$\displaystyle y_h(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)] $$ The figure below shows a plot of the solution.
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Part B

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$$\displaystyle \lambda_{1,2}=-3 $$     (9.2) 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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$$\displaystyle \lambda^{2}+6\lambda+9=0 $$ Corresponding Homogeneous ODE:
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$$\displaystyle y''+6y'+9y=0 $$ 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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$$\displaystyle y_h(x)=c_1e^{\lambda x}+c_2xe^{\lambda x} $$ Applying the root value to the form above, we obtain the general solution:
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$$\displaystyle y_h(x)=c_1e^{-3x}+c_2xe^{-3x} $$ 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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$$\displaystyle c_1=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=-3c_1e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x} $$ 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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$$\displaystyle c_2=3 $$ 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) 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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$$\displaystyle \lambda^{2}-3\lambda+-10=0 $$ Corresponding Homogeneous ODE:
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$$\displaystyle y''-3y'-10y=0 $$ 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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$$\displaystyle y_h(x)=c_1e^{\lambda _1 x}+c_2e^{\lambda_2 x} $$ Applying the distinct root values to the form above, we obtain the general solution:
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$$\displaystyle y_h(x)=c_1e^{-2 x}+c_2e^{5 x} $$ 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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$$\displaystyle c_1+c_2=1 $$ We now take the first derivative of the general solution:
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$$\displaystyle {y_h}'(x)=-2c_1e^{-2x}+5c_2e^{5x} $$ Applying the initial condition $$\ y'(0)=0 $$:
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$$\displaystyle 0=-2c_1e^{0}+5c_2e^{0} $$
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$$\displaystyle 0=-2c_1+5c_2 $$ 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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$$\displaystyle c_1=\frac{5}{7} $$ 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.