# Thread: Solving a differential equation using Laplace transform

1. ## Solving a differential equation using Laplace transform

Hi there. I have to solve the differential equation using the laplace transform.

$ty''+2y'+ty=0,y(0)=1,y(\pi)=0$

So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

$-\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0$

$-s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy$

Then: $y(s)=-\arctan(s)+C$

Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.

2. ## Re: Solving a differential equation using Laplace transform

Originally Posted by Ulysses
Hi there. I have to solve the differential equation using the laplace transform.

$ty''+2y'+ty=0,y(0)=1,y(\pi)=0$

So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

$-\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0$

$-s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy$

Then: $y(s)=-\arctan(s)+C$

Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.
Yes you have definitely done something wrong. When you take the Laplace Transform, you should get some function of s, not a differential equation that involves s.

3. ## Re: Solving a differential equation using Laplace transform

Thats because the differential equation has variable coefficients, so one gets a differential equation of s. I've seen other exercises likes this. Perhaps I wasn't clear, the function I'm looking is y(t), y(s) denotes it's Laplace transform.

When I apply the laplace transform on the equation I get:
$\mathcal{L} \left [ ty'' \right ] +2 \mathcal{L} \left [ y' \right ]+ \mathcal{L} \left [ ty \right ]=0$

Then using $\mathcal{L} \left [t^n F(t)\right ]=(-1)^n\frac{d^n f}{ds^n}(s)$
And $\mathcal{L} \left [ F^n(t) \right ]=s^n \mathcal{L} \left [ F(t)\right ]-s^{n-1}F(0)-s^{n-2}F'(0)-...-F^{n-1}(0)$

I get for example:
$\mathcal{L} \left [ ty'' \right ]=-\frac{d}{ds}(s^2 \mathcal{L}\left [ y \right ]-sy(0)-y'(0))$
I don't know y'(0), but its a constant, then when deriving I get to the second equation that I put at the beginning.

4. ## Re: Solving a differential equation using Laplace transform

Originally Posted by Ulysses
Hi there. I have to solve the differential equation using the laplace transform.

$ty''+2y'+ty=0,y(0)=1,y(\pi)=0$

So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

$-\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0$

$-s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy$

Then: $y(s)=-\arctan(s)+C$

Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.
You got $\frac{dF}{ds} = \frac{-1}{s^2 + 1}$ which is correct.

Now note that $LT[t f(t)] = - \frac{dF}{ds}$ and so you have $LT[t f(t)] = \frac{1}{s^2 + 1} \Rightarrow t f(t) = ....$