# Solving the IVP ty′′-ty′+y=1 where y(0)=0 and y′(0)=2

Solving the IVP
$t{y}^{″}-t{y}^{\prime }+y=1$
where $y\left(0\right)=0$ and ${y}^{\prime }\left(0\right)=2$
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grizintimbp
$t{y}^{″}-t{y}^{\prime }+y=1$
$-{\left({s}^{2}\mathcal{L}\left(y\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)}^{\prime }+{\left(s\mathcal{L}\left(y\right)-y\left(0\right)\right)}^{\prime }+\mathcal{L}\left(y\right)=\frac{1}{s}$
$\left(s-{s}^{2}\right){\mathcal{L}}^{\prime }\left(y\right)+\left(2-2s\right)\mathcal{L}\left(y\right)=\frac{1}{s}$
${\mathcal{L}}^{\prime }\left(y\right)+\frac{2}{s}\mathcal{L}\left(y\right)=\frac{1}{{s}^{2}\left(1-s\right)}$
with integrating factor ${s}^{2}$ we have
$\begin{array}{rl}\left({s}^{2}\mathcal{L}\left(y\right){\right)}^{\prime }& =\frac{1}{1-s}\\ y& ={\mathcal{L}}^{-1}\frac{-\mathrm{ln}\left(1-s\right)}{{s}^{2}}+Ct\\ & ={\int }_{0}^{t}\frac{{e}^{x}}{x}\left(t-x\right)dx+Ct\\ & =t{\int }_{0}^{t}\frac{{e}^{x}}{x}dx-{e}^{t}+1+Ct\\ & =t\mathrm{Ei}\left(t\right)-{e}^{t}+1+Ct\end{array}$