Question

procedure for laplace laplace y''-4y'+4y=t^3 e^(2t)

Answer 1

Using the definition, or do you have some sort of table to common transforms?

Answer 2

Also, you won't be able to find an exact solution, since you didn't provide the initial values. \(\big(y(0)=\cdots,~y'(0)=\cdots\big)\)

Answer 3

sorry the initial values are y(o)=0 and y'(0)=0

Answer 4

I have \[s ^{2} Y(s)-sy'(0)-y''(0)-4[sY(s)-y(0)]+4Y(s)\]

Answer 5

=?????

Answer 6

Ok. Take the Laplace transform of both sides, you should have
\[\bigg[s^2Y(s)-\color{red}{s~y(0)-y'(0)}\bigg]-4\bigg[s~Y(s)-y(0)\bigg]+4Y(s)=\int_0^\infty t^3e^{2t}e^{-st}~dt\]

Answer 7

\[\left(s^2-4s+4\right)Y(s)=\int_0^\infty t^3e^{2t}e^{-st}~dt\]
The reason I asked earlier whether you had a table of transforms is because doing the integration by hand is rather tedious, since it involves multiple steps of integrating by parts.

Answer 8

that's right. Let me look for the integral solution....

Answer 9

I remember one of the transforms was the following: If you're given a function of the form \(e^{at}~f(t)\), then its transform is \(F(s-a)\). In other words, the transform will be whatever the transform of \(f(t)\) is, but with a shift of \(a\).

In this case, \(f(t)=t^3\), whose transform is \(\dfrac{3!}{s^{3+1}}=\dfrac{6}{s^4}\). We also have \(a=2\), so there's a shift of 2. So, the transform of the right side of the equation above is
\[\mathscr{L}\left\{t^3e^{2t}\right\}=\frac{6}{(s-2)^4}\]

Answer 10

\[\begin{align*}Y(s)&=\frac{6}{(s-2)^4(s^2-4s+4)}\\
&=\frac{6}{(s-2)^4(s-2)^2}
\end{align*}\]

Answer 11

that's right

Answer 12

Do you know where to go from here? How to find the inverse transform
\[\mathscr{L}^{-1}\left\{\frac{6}{(s-2)^6}\right\}~?\]

Answer 13

but any inverse laplace look like that
or it will be better to factorize??
(Sorry, for my ortography)

Answer 14

Think back to the transform of \(f(t)=t^n\). Its transform is \(\dfrac{n!}{s^{n+1}}\).

The degree of the denominator is 6, so you're looking for some function of degree 5 (so that n=5). Therefore, in the numerator you should have 5!. What can you multiply by to make it right?

Answer 15

And again, the \(s-2\) in the denominator indicates a shift, so the inverse transform will be something multiplied by \(e^{2t}\).

Answer 16

ok, let me sovle it and I will show you :)

Answer 17

\[6/5! t^5\]