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Solve the differential equation with the laplace transformation. $$y''-4y'+9y=9\quad,\quad y(0)=0 \quad ,\quad y'(0)=-8$$

I will solve this question to this state, but I cannot continue.

\begin{align} s^2 y(s)-4sy(s)+9y(s)=\frac{9}{s} \end{align}

\begin{align} y(s)&=\frac{9}{s}(\frac{1}{s^2}-4s+9) \\ &=\frac{1}{s}-\frac{s-4}{(s-2)^2}+5 \\ &=\frac{1}{s}-\frac{s-2}{(s-2)^2}-5 + \frac{2}{(s-2)^2}+5 \end{align} I did this so far, but I cannot transform last term.

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Hello and welcome to math.stackexchange.com. Since you are new here, I'd like to point out two things: 1. People don't like being addressed with imperatives here; please reformulate your question into an actual question format. 2. It's considered polite to describe not just the problem you are trying to solve, but also what you have tried and where you are stuck. This is particularly true if the question you ask stems from a homework problem (in which case it is also recommended to use the "homework" tag). About the question: Do you have a table of standard Laplace transforms available? –  Johannes Kloos Mar 2 '13 at 19:57
    
Sorry Joannes, i'm very new at this. :) I corrected the question. :) yes i have a table of standart Laplace transforms. but i cannot transform last term. –  osmantan Mar 2 '13 at 20:14
    
@glacier I wonder whether you have seen this thread on meta: meta.math.stackexchange.com/questions/13289/… –  Martin Sleziak Apr 6 at 4:54

1 Answer 1

You should first know that, if

$$\hat{y}(s) = \int_0^{\infty} dt \: y(t) e^{-s t}$$

is the LT of $y$, then the LT of the first and second derivative of $y$ is

$$\hat{y'}(s) = -y(0) + s \hat{y}(s) = s \hat{y}(s)$$ $$\hat{y''}(s) = -y'(0) - s y(0) + s^2 \hat{y}(s) = 8 + s^2 \hat{y}(s)$$

which may be shown by integration by parts. I think you erred here in not including the initial conditions. The differential equation becomes

$$[8 + s^2 \hat{y}(s)] -4 [ s \hat{y}(s)] + 9\hat{y}(s) = 9 \int_0^{\infty} dt \: e^{-s t} = \frac{9}{s} $$

or

$$(s^2-4 s+9) \hat{y}(s) = \frac{9}{s}-8 \implies \hat{y}(s) = \frac{9-8 s}{s (s^2-4 s+9)}$$

To find the inverse transform, use a partial fraction decomposition. Note that

$$s^2-4 s+9=0 \implies s_{\pm} = 2 \pm i \sqrt{5}$$

I will spare you further algebra; the decomposition is

$$\frac{9-8 s}{s (s^2-4 s+9)} = \frac{1}{s} + \frac{B}{s-s_+} + \frac{C}{s-s_-}$$

where $B = (4+s_+)/(s_+-s_-) = -\frac{1}{2} - i \frac{3}{\sqrt{5}}$ and $C= -(4+s_-)/(s_+-s_-) = -\frac{1}{2} + i \frac{3}{\sqrt{5}}$.

At this point, you should be able to do the inverse transform by lookup. After some simplification, the result is

$$y(t) = 1 + e^{2 t} \left ( -\cos{\sqrt{5} t} + \frac{6}{\sqrt{5}} \sin{\sqrt{5} t} \right )$$

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