Laplace inverse problem

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Jun 26th 2008, 08:30 AM
tttcomrader

Laplace inverse problem

Transform the differential equation to find a nontrivial solution such that x(0) = 0.

$tx''+(t-2)x'+x=0$

What I have so far after the laplace transform is:

$-2sX-s^2X'-X'-2X-2sX+X=0$
$\frac {X'}{X} = - \frac {4s +1}{s^2+1}$

Then taking the integral, I have $lnX = -2ln(s^2+1) - tan^{-1}s$
Jun 26th 2008, 01:27 PM
bleesdan

Try e to the (what you had)
The X is found, and then you backtrack to x.
Jun 26th 2008, 04:03 PM
mr fantastic

Quote:

Originally Posted by tttcomrader

Transform the differential equation to find a nontrivial solution such that x(0) = 0.

$tx''+(t-2)x'+x=0$

What I have so far after the laplace transform is:

$-2sX-s^2X'-X'-2X-2sX+X=0$
$\frac {X'}{X} = - \frac {4s +1}{s^2+1}$

Then taking the integral, I have $lnX = -2ln(s^2+1) - tan^{-1}s$

I have no time now but will take a closer look later.
Jun 26th 2008, 05:27 PM
topsquark

Quote:

Originally Posted by mr fantastic

I have no time now but will take a closer look later.

Just for the record, Mathematica doesn't like either as a closed form solution. (That doesn't always mean that there isn't one, though.)

-Dan
Jun 26th 2008, 07:43 PM
NonCommAlg

Quote:

Originally Posted by tttcomrader

Transform the differential equation to find a nontrivial solution such that x(0) = 0.

$tx''+(t-2)x'+x=0$

What I have so far after the laplace transform is:

$-2sX-s^2X'-X'-2X-2sX+X=0$
$\frac {X'}{X} = - \frac {4s +1}{s^2+1}$

Then taking the integral, I have $lnX = -2ln(s^2+1) - tan^{-1}s$

what you got is incorrect! if, as your notation, $\mathcal{L}(x(t))=X(s),$ then since we are given that $x(0)=0,$ we will get:

$\mathcal{L}(tx''(t))=-s^2 X'(s) - 2sX(s), \ \ \ \mathcal{L}(tx'(t))=-sX'(s)-X(s), \ \ \ \mathcal{L}(x'(t))=sX(s).$ thus the transformation of your

differential equation is: $-(s^2+s)X'(s) - 4sX(s)=0.$ that will give you $X(s)=\frac{C}{(s+1)^4},$ where $C$ is any constant.

so the general solution is: $x(t)=kt^3e^{-t},$ where $k$ is any constant. Q.E.D.