Solving ODE by Laplace transformation

Question

Consider the differential equation $ty''(t)+2y'(t)+ty=0$, $t>0$, $y(0+)=1$ and $y'(0+)=0$. If $Y(s)$ is the Laplace transform of $y(t)$ then find the value of $Y(1)$.

Taking Laplace transform both sides of the given equation and solving then we get, $$Y(s)=-\tan^{-1}(s)+C$$ where $C$ is integrating constant.

My question is how to find the constant $C$ ? Any hint. please.

Answer 1

Taking Laplace transform both sides of the given equation and solving then we get: $$Y(s)=-\tan^{-1}(s)+C$$

We are going to assume that whatever our solution is, it is of exponential order. This means that:

$$\underset{s\to \infty }{\text{lim}}\left(-\tan ^{-1}(s)+C\right)=0$$

The first term does go to $-\frac{\pi }{2}$ in the limit. Therefore, we must have C = $\frac{\pi }{2}$ in order for this to be the transform of our solution.

So, the transform of our solution, as well as the solution is:

$$\mathcal{L}_s^{-1}\left[-\tan ^{-1}(s)+\frac{\pi }{2}\right](t)=y(t)=\frac{\sin (t)}{t}$$ You can find similar examples from book on page:222-224.

Answer 2

Hint.

An easy way to solve the ODE

$$ (t y(t))'' + t y(t) = 0\ \Rightarrow t y(t) = c_1\cos t+c_2\sin t $$

etc.