given the ode:$t^{2}u''+tu'+(t^{2}-1)u=0;u(0)=1,u'(0)=0$
solve with Laplace transform.
My try: $t^{2}u''+tu'+(t^{2}-1)u=0\\Lu''=s^{2}Lu-1\\Lu'=sLu-1\\t^{2}(s^{2}Lu-1)+t(sLu-1)+(t^{2}-1)Lu=0\\t^{2}s^{2}Lu-t^{2}+tsLu-t+(t^{2}-1)Lu=0\\t^{2}s^{2}Lu+tsLu+(t^{2}-1)Lu=t^{2}+t\\Lu(t^{2}s^{2}+ts+t^{2}-1)=t^{2}+t\\Lu=\frac{t^{2}+t}{(t^{2}s^{2}+ts+t^{2}-1)}\\Lu=\frac{1}{1+s^{2}}+\frac{1+t-st+s^{2}t}{(1+s^{2})(-1+st+t^{2}+s^{2}t^{2})}$
which seems to be a dead end. Edit: I was able to find L(u)
$\begin{array}{c} t^{2}u''+tu'+(t^{2}-1)u=0\\ Lt^{2}u''+Ltu'+Lt^{2}\text{u}-Lu=0\\ Lt^{2}u''+Ltu'+Lt^{2}\text{u}-Lu=0\\ (Lu'')''-(Lu')'+(Lu)''-Lu=0\\ u(0)=1,u'(0)=0\\ (Lu'')''-(Lu')'+(Lu)''-Lu=0\\ (s^{2}Lu-1)''-(sLu-1)'+(Lu)''-Lu=0\\ (s^{2}L''u+4sL'+2Lu)-(sL'u+Lu)+(L''u)-Lu=0\\ s^{2}L''u+4sL'+2Lu-sL'u-Lu+L''u-Lu=0\\ (s^{2}+1)L''u+3sL'=0\\ (x^{2}+1)\text{y''}+3xy'=0\\ y''+\frac{3x}{(x^{2}+1)}y'=0\\ z'+\frac{3x}{(x^{2}+1)}z=0\\ \frac{z'}{z}=-\frac{3x}{(x^{2}+1)}\\ \log|z|=-\frac{3}{2}log(1+x^{2})+c\\ z=c(1+x^{2})^{-\frac{3}{2}}\\ y'=c(1+x^{2})^{-\frac{3}{2}}\\ Lu=y=c_{1}+\frac{c_{2}x}{\sqrt{x^{2}+1}} \end{array}$
$$t^{2}u''+tu'+(t^{2}-1)u=0$$ $$u(0)=1,u'(0)=0$$ You can't keep the $t$ variable once you have applied the Laplace transform.
Laplace transform of $t^2u''$ is: $$L t^2u''=\dfrac {d^2}{ds^2}Lu''=\dfrac {d^2}{ds^2}(s^2U(s))-su'(0)-u(0))$$ $$L t^2u''=\dfrac {d^2}{ds^2}(s^2U(s))-su'(0))$$ $$L t^2u''=\dfrac {d^2}{ds^2}(s^2U(s))$$ And: $$L tu'=-\dfrac {d}{ds}Lu'=-\dfrac {d}{ds}(sU(s)-u(0))$$ $$=-\dfrac {d}{ds}(sU(s))=-(U(s)+sU'(s))$$ The Lapalce transform of the last term of the ifferential equation: $$L(t^2-1)u=-U(s)+\dfrac {d^2}{ds^2}(U(s))$$ $$=-U(s)+U''(s)$$
Note that the DE is Bessel's differential equation of order $1$. Not that easy to solve with Laplace Transform. Take a look here