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Solve each of the following differential equations with initial values using the Laplace Transform.

$(b)\space y''-4y'+4y=0$ Where $y(0)=0$ and $y'(0)=3$

What I have so far:

$p^2L[y]-3-4pL[y]+4L[y]=0$

$L[y]=\frac{3}{p^2-4p+4}=\frac{3}{(p-2)^2}$ I'm not sure where to go from here..

$(c)\space y''+2y'+2y=2$ Where $y(0)=0$ and $y'(0)=1$

What I have so far:

$p^2L[y]-1+2pL[y]+2L[y]=L[2]$

$L[y](p^2+2p+2)=\frac{2+p}{p}$

$L[y]=\frac{2+p}{p((p+1)^2+1)}$ From here I tried using partial fractions:

$\frac{A}{p}+\frac{B}{(p+1)^2+1}$ I found A=1 and B=-1. I'm fairly sure that is correct, but I'm not sure where to go from here.

$(d)\space y''+y'=3x^2$ Where $y(0)=0$ and $y'(0)=1$

What I have so far:

$p^2L[y]-1+pL[y]=L[3x^2]=\frac{6}{p^3}$

$L[y]=\frac{6}{p^4(p+1)}$

$(e)y''+2y'+5y=3x^{-x}sin(x)$ Where $y(0)=0$ and $y'(0)=3$

What I have so far:

$p^2L[y]-3+2pL[y]+5L[y]=\frac{3}{(p+1)^2+1}$

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    I have a table that gives me standard Laplace transforms, such as $sin(ax)$, but i'm not sure how to implement this.. Could you finish $(c)$ or $(d)$ for me? So, I can try solving the others? Thanks2012-11-02

1 Answers 1

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For the question c:

$ L\{y\}=\frac{1}{p}-\frac{1}{(p+1)^2+1} $

The inverse Laplace becomes,

$ y(x)=1-e^{-x}\sin(x) $

Explaination:

$ L^{-1}\{\frac{1}{p}\}=1 $

$ L^{-1}\{\frac{1}{p^2+1^2}\}=\sin(x) $

$ L^{-1}\{\frac{1}{(p-(-1))^2+1^2}\}=e^{-x}\sin(x) $

For the question e:

$ \frac{L\{y\}}{3}=\frac{1}{(p+1)^2+2^2}+\frac{1}{(p+1)^2+2^2}\frac{1}{(p+1)^2+1^2} $

$ \frac{y(x)}{3}=e^{-x}\sin(2x)+(e^{-x}\sin(2x))*(e^{-x}\sin(x)) $

$ \frac{y(x)}{3}=e^{-x}\sin(2x)+\int_0^x\left[e^{-\lambda}\sin(2\lambda) e^{-(x-\lambda)}\sin(x-\lambda)\right]\textrm{d}\lambda $

$ y(x)=2e^{-x}\left(\sin x+\sin(2x)\right) $

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    The final answer for question number e must be checked again. I feel there is a tiny mistake there.2012-11-02