I'm stuck solving this equation:$$t^2y''+2ty'+t^2y=0 \qquad \text{with} \quad y(0)=0 , \; y'(0)=2 $$
I applied the Laplace transform and get this:
$$s^2Y''(s)+Y''(s) + 2sY'(s)=0$$
I tried to solve this by Series, is it correct, or is there another way to solve it? (the answer is $y=-C\frac{\sin t}{t}$)
It is correct that $y=-C\frac{\sin(t)}{t}$ is a particular family of solutions of the ODE : $$t^2y''+2ty'+t^2y=0 \qquad\text{without condition.} $$ Then, $y(0)=-C$ . Hense, with condition $y(0)=0 \quad\to\quad C=0 \quad\to\quad y(x)=0$
With this solution we have : $y'(0)=0$ , hence $y'(0)$ cannot be $=2$.
As a consequence, the ODE with the two conditions $y(0)=0$ and $y'(0)=0$ has no solution.
An alternative way than solving with series is the change of function $y(t)=\frac{y(t)}{t}$ which leads to : $$f''+f=0\quad\to\quad f(x)=c_1\sin(t)+c_2\cos(t)$$ The general solution of the ODE (without condition) is : $$y(t)=c_1\frac{\sin(t)}{t}+c_2\frac{\cos(t)}{t}$$ The condition $y(0)=0$ implies $c_1=0$ and $c_2=0 \quad\to\quad y(t)=0$
This is consistent with the above result. Of course, the conclusion is the same : With the second condition $y'(0)=2$, there is no solution.