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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:


$L[y]=\frac{3}{p^2-4p+4}=\frac{3}{(p-2)^2}$ 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:



$L[y]=\frac{2+p}{p((p+1)^2+1)}$ 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:



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

What I have so far:


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You need one step further, find the inverse laplace! – kiss my armpit Nov 2 '12 at 15:01
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? Thanks – Dmitri.Mendeleev Nov 2 '12 at 15:04
up vote 1 down vote accepted

For the question c:

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

The inverse Laplace becomes,

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


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

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

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

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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So it is a rule that for $-\frac{1}{(p+1)^2+1}$ the inverse laplace is $-e^xsin(x)$? – Dmitri.Mendeleev Nov 2 '12 at 15:13
I got them all now, except for $(e)$. Could you help me finish that one? Thanks – Dmitri.Mendeleev Nov 2 '12 at 15:53
The final answer for question number e must be checked again. I feel there is a tiny mistake there. – kiss my armpit Nov 2 '12 at 18:25

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