How do i find the general solution of this equation using laplace transform?? its a practice problem for my exam but im not sure how to solve it.

y'''-4y'=0 when y(0)=1, y'(0)=2, y(0)''=-20

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- Nov 7th 2010, 08:29 AMJuggalomikeLaplace transform
How do i find the general solution of this equation using laplace transform?? its a practice problem for my exam but im not sure how to solve it.

y'''-4y'=0 when y(0)=1, y'(0)=2, y(0)''=-20 - Nov 7th 2010, 09:59 AMTheEmptySet
The laplace transform of

$\displaystyle \frac{d^ny}{dt^n}=s^nY-\sum_{k=0}^{n-1}s^{n-k-1}y^{k}(0)$

So in your case

$\displaystyle \mathcal{L}\{y'''\}=s^3Y-s^2y(0)-sy'(0)-y''(0)=s^3Y-s^2-2s+20$

and

$\displaystyle \mathcal{L}(y')=sY-1$

This gives the algebraic equation

$\displaystyle s^3Y-s^2-2s+20-4(sY-1)=0$

Now just solve for Y and invert the transform - Nov 7th 2010, 10:16 AMJuggalomike
ah great, thanks alot, was doing the y''' wrong.