Solving the IVP $ty''-ty'+y=1$ where $y(0) = 0$ and $y'(0)=2$.

Question

Solving the IVP $$ty''-ty'+y=1$$ where $y(0) = 0$ and $y'(0)=2$.

I'm struggling to get started. I've solved some problems of similar fashion, but without the t factor on the left hand side.

Answer 1

$$ty''-ty'+y=1$$ $$-\left(s^2{\cal L}(y)-sy(0)-y'(0)\right)'+\left(s{\cal L}(y)-y(0)\right)'+{\cal L}(y)=\dfrac{1}{s}$$ $$(s-s^2){\cal L}'(y)+(2-2s){\cal L}(y)=\dfrac{1}{s}$$ $${\cal L}'(y)+\dfrac2s{\cal L}(y)=\dfrac{1}{s^2(1-s)}$$ with integrating factor $s^2$ we have \begin{align} (s^2{\cal L}(y))' &= \dfrac{1}{1-s} \\ y&={\cal L}^{-1}\dfrac{-\ln(1-s)}{s^2}+Ct \\ &=\int_0^t\dfrac{e^x}{x}(t-x)dx+Ct \\ &= t\int_0^t\dfrac{e^x}{x}dx-e^t+1+Ct \\ &= t\operatorname{Ei}(t)-e^t+1+Ct \end{align}