Past exam of the mathematical methods I course of the University of Oxford / 2021 resit / 11

On an interval with $x\ne 0$, this is a homogeneous first-order differential equation. Put $y=xv(x)$, so $y^{\prime }=v+x{v}^{\prime }$. Substitution and division by $x^2$ giveThe constant choices $v=0$ and $v=2$ give $y=0$ and $y=2x$. Otherwise the separable differential equation givesAbsorbing the sign and the exponential integration constant yields $(v-2)/v=C{x}^2$. Relabelling the arbitrary constant then givesThe first family already includes $y=2x$ at $K=0$. Choose any interval where its denominator is nonzero. These give all solutions on intervals avoiding $x=0$. Both families extend through $x=0$ with $y(0)=0$, as required by the original ordinary differential equation. Because it is singular there, differentiable continuations of the nonzero family may use different $K$ on the two sides: both one-sided derivatives at zero equal $2$.

For real solutions work on $x>-1$. Dividing by $2(1+x)$ puts the linear ordinary differential equation into the formIts integrating factor isThe product rule therefore givesTwice using integration by parts, or differentiating the proposed antiderivative, gives $\int x^2{e}^{-x}dx=-e^{-x}(x^2+2x+2)$. HenceThe endpoint $x=-1$ is singular. A finite continuous extension would require $C=e/2$, but then, with $h=x+1\downarrow 0$, $y=\frac{1}{2}\sqrt{h}+O(h^{3/2})$. Its derivative diverges, so even this choice has no differentiable extension to that endpoint.

The zero function is a solution on any interval avoiding $x=0$. For a nonzero solution, the Bernoulli differential equation substitution $u=1/y$ gives $u^{\prime }=-y^{\prime }/y^2$ andAn integrating factor is $1/x$ on either side of zero. ThusThe full solution set isUse connected intervals with $x\ne 0$ and, in the nonzero family, $Cx-x^2\ne 0$. The separate zero solution would be lost by dividing by $y^2$.