= Solution
The <parameter sensitivity equation> follows by differentiating with respect to $\mu$ and writing $u=y_\mu$:
$$
u_x=u+x+y^2+2\mu yu,
\qquad u(0,\mu)=0.
$$
At $\mu=0$, $y_x=y$, $y(0)=1$, so $y(x,0)=e^x$. Then
$$
u_x-u=x+e^{2x},\qquad u(0,0)=0,
$$
and the integrating factor $e^{-x}$ yields
$$
u(x,0)=e^{2x}-x-1.
$$
Equating powers in $y=y_0+\mu y_1+\mu^2y_2+\cdots$ gives
$$
y_0=e^x,\qquad y_1=e^{2x}-x-1,
$$
$$
y_2'-y_2=2y_0y_1,\qquad y_2(0)=0.
$$
Consequently
$$
\boxed{y_2=e^x(e^{2x}-1-x^2-2x).}
$$
Back to article page
