# Inverse Laplace

• November 1st 2008, 02:18 AM
Inverse Laplace
Well i got this question from my tutorial. Given the answers as well with it.Still couldn't understand how my lecturer got the answer.

Using Laplace Transform methods, solve the following ode:

$tV'' + 2V' + tV = 0$

subject to the following boundary conditions:

$V(0) = 1, V( \pi ) = 1$

Hint: You may assume that $L^{-1} \{ tan^{-1} \left( \frac{1}{s} \right) \} = \frac {sin t}{t}$

This is what i did:

$- \frac{d}{ds} \{ s^2V(s) - sv(0) - v'(0) \} + 2\{ sV(s) - v(0) \} - \frac{d}{ds}V(s) = 0$

$V'(s)(-s^2 - 1) - v(0) = 0$

$V(s) = \int \frac{1}{s^2 +1} ds$

letting $u = \frac{1}{s}$

$V(s) = \int \frac{1}{1+u^2} du$

$V(s) = tan^{-1} \frac{1}{s} + C$

Then by inversing laplace

$V(t) = \frac{sin t}{t} +L^{-1} C$

the answer given was $V(t) = \frac{sint}{t} + \frac{\pi}{t}$

i don't know how did $L^{-1} C$ becomes $\frac{\pi}{t}$
• November 1st 2008, 03:41 AM
mr fantastic
Quote:

Well i got this question from my tutorial. Given the answers as well with it.Still couldn't understand how my lecturer got the answer.

Using Laplace Transform methods, solve the following ode:

$tV'' + 2V' + tV = 0$

subject to the following boundary conditions:

$V(0) = 1, V( \pi ) = 1$

Hint: You may assume that $L^{-1} \{ tan^{-1} \left( \frac{1}{s} \right) \} = \frac {sin t}{t}$

This is what i did:

$- \frac{d}{ds} \{ s^2V(s) - sv(0) - v'(0) \} + 2\{ sV(s) - v(0) \} - \frac{d}{ds}V(s) = 0$

$V'(s)(-s^2 - 1) - v(0) = 0$

$V(s) = \int \frac{1}{s^2 +1} ds$

letting $u = \frac{1}{s}$

$V(s) = \int \frac{1}{1+u^2} du$

$V(s) = tan^{-1} \frac{1}{s} + C$

Then by inversing laplace

$V(t) = \frac{sin t}{t} +L^{-1} C$

the answer given was $V(t) = \frac{sint}{t} + \frac{\pi}{t}$

i don't know how did $L^{-1} C$ becomes $\frac{\pi}{t}$

The answer given is wrong as far as I can see.