Consider the differenial equation $$t\frac{d^2y}{dt^2}+2\frac{dy}{dt}+ty=0\tag{1}$$ $$t>0,y(0+)=1,y'(0+)=0$$ If $Y(s)$ is Laplace transform of $y(t)$, then find the value of $Y(1)$.
My attempt: we know that $\mathcal{L}({tf(t)})=-\frac{d}{ds}F(s)$,where $F(s)$ is the Laplace transform of $f(t)$. Now taking Laplace transform both side in equation $(1)$, we have $$\begin{align*}&\mathcal{L}[t\frac{d^2y}{dt^2}+2\frac{dy}{dt}+ty]=0 \\\Rightarrow&\mathcal{L}[t\frac{d^2y}{dt^2}]+2\mathcal{L}[\frac{dy}{dt}]+\mathcal{L}[ty]=0 \\\Rightarrow&-\frac{d}{ds}[s^2Y(s)-sy'(0)-y(0)]+2[sY(s)-y(0)]-\frac{d}{ds}Y(s)=0 \\\Rightarrow&-2sY(s)-s^2Y'(s)+2sY(s)-2-Y'(s)=0 \\\Rightarrow&Y'(s)=\frac{-2}{1+s^2} \\\Rightarrow&Y(s)=-2\tan^{-1}s+C\end{align*}$$
Where $C$ is some constant, therefore $Y(1)=-2\tan^{-1}(1)+C=-\frac{\pi}{2}+C$.
What I need to know is only constant $C$. I tried my best but didn't found the value of this constant.
Any help will be appreciable
Thanks in advance.
I spotted one mistake: $\mathcal L[y'']=s^2Y(s)-s\color{red}y(0)-\color{red}{y'}(0)$. Using the correct formula you should get $Y(s)=C-\tan^{-1}s$. To find $C$, simply use the fact that $\lim_{s\to\infty}Y(s)=0$.