How do I solve with Laplace Transform $t^{2}y''-2ty'+2y=0$

Question

I have to solve $t^{2}y''-2ty'+2y=0$ with the initial conditions $y(0)=0$, $y'(0)=1$ using the Laplace transformation. I only could do this

$s^{2}\mathcal{Y}''+6s\mathcal{Y}'+6\mathcal{Y}=0$

where $\mathcal{Y}$ is the Laplace transformation of the function $y(t)$. What can I do?

Answer 1

$$t^{2}y''-2ty'+2y=0$$ $$\dfrac {d^2}{ds^2}\mathcal {L} \{y''\}+2\dfrac {d}{ds}\mathcal {L} \{y'\}+2Y(s)=0$$ $$\dfrac {d^2}{ds^2} \{s^2Y(s)\}+2\dfrac {d}{ds} \{sY(s)\}+2Y(s)=0$$ $$\dfrac {d^2}{ds^2} \{s^2Y(s)\}+2sY'(s)+4Y(s)=0$$ $$s^2Y''(s)+6sY'(s)+6Y(s)=0$$ This is Cauchy-Euler's differential equation .

Try $Y(s)=s^m$.

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Another way:

$$s^2Y''(s)+6sY'(s)+6Y(s)=0$$ Multiply by $\mu (s)=s$ $$s^3Y''(s)+6s^2Y'(s)+6sY(s)=0$$ You can rewrite it as: $$(s^3Y'(s))'+(3s^2Y(s))'=0$$ $$(s^3Y(s))''=0$$ Integrate twice. Apply inverse Laplace transform and initial conditions.