 Solution

Write $y=e^{-px}v$. Direct differentiation givesThe homogeneous linear differential equation therefore has $v=A+Bx$. A linearly independent pair of solutions isTheir Wronskian is $e^{-2px}$, which never vanishes; the argument also includes $p=0$.

For the inhomogeneous linear differential equation, the same substitution gives $v^{\prime \prime }=1$. Since $y(0)=0$ implies $v(0)=0$, and $y^{\prime }(0)=0$ then implies $v^{\prime }(0)=0$, integration gives $v=x^2/2$. Thus the required solution of the initial value problem is