Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 10

The integrating factor is $e^{2x}$, giving $(e^{2x}y{)}^{\prime }=1+e^{2x}(x^2+2x)$. A quadratic particular solution for the polynomial forcing is $x^2/2+x/2-1/4$, while the resonant exponential forcing contributes $x{e}^{-2x}$. ConsequentlySubstitution gives $y^{\prime }+2y=e^{-2x}+x^2+2x$, and the free multiple of $e^{-2x}$ is the homogeneous solution.

Integrate the equation without dividing by $y$:This holds on a connected interval excluding $x=-1$. The real solutions are differentiable branches of this implicit curve. Where the radicand is positive, $y=\pm \sqrt{2x-2\log |x+1|+C}$, with a locally fixed sign.

A zero of $y$ can belong to a classical solution only at $x=0$, since the original equation then requires $x/(x+1)=0$. For $C=0$ near zero, $2x-2\log (1+x)=x^2+O(x^3)$; the smooth continuations have slopes $+1$ or $-1$ and switch between square-root signs across zero. Keeping the principal nonnegative square root across zero instead gives a cusp. Thus the implicit expression requires a smooth branch choice, not an unrestricted global choice of one square-root sign.

The operator is $(D+1{)}^2$. Set $y=e^{-x}u$; direct differentiation gives $(D+1{)}^{2}y=e^{-x}{u}^{\prime \prime }$. Hence $u^{\prime \prime }=3x^2$, and two integrations giveThe two free constants multiply the independent homogeneous solutions $e^{-x}$ and $x{e}^{-x}$.