= Solution
Write $y=e^{-px}v$. Direct differentiation gives
$$
y''+2py'+p^2y=e^{-px}v''.
$$
The <homogeneous linear differential equation> therefore has $v=A+Bx$. A <linearly independent> pair of solutions is
$$
\boxed{y_1=e^{-px},\qquad y_2=xe^{-px}.}
$$
Their <Wronskian> is $e^{-2px}$, which never vanishes; the argument also includes $p=0$.
For the <inhomogeneous linear differential equation>, the same substitution gives $v''=1$. Since $y(0)=0$ implies $v(0)=0$, and $y'(0)=0$ then implies $v'(0)=0$, integration gives $v=x^2/2$. Thus the required solution of the <initial value problem> is
$$
\boxed{y(x)=\frac{x^2}{2}e^{-px}.}
$$
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