Past exam of the mathematical methods I course of the University of Oxford / 2013 / 12

The second equation gives $y=t+x^{\prime }-x$. Differentiating it gives $y^{\prime }=1+x^{\prime \prime }-x^{\prime }$. Substitution into the first equation cancels the $x^{\prime }$ terms and the constant forcing, leaving $x^{\prime \prime }+x=0$. HenceThis has two independent constants, as expected for the first-order linear system of differential equations. At zero, the conditions give $A=0$ and $B=1$. Thus the required particular solution isChecking directly, $x^{\prime }+y^{\prime }+x=\cos t+(1-\sin t-\cos t)+\sin t=1$, while $-x^{\prime }+x+y=t$. This also verifies the signs introduced during elimination.

On an interval with $x\ne -1$, an integrating factor is $(x+1{)}^{-2}$. Multiplication givesIntegration therefore yieldsThe polynomial formula has a smooth continuation at $x=-1$, but the original coefficient $2/(x+1)$ is undefined there. Solution intervals for the stated equation must avoid that point, and the constants on its two sides need not agree.

The logarithm requires $x>0$ for real-valued solutions. Put $v=xy$. Since $v^{\prime }=x{y}^{\prime }+y$, the equation becomes $v^{\prime }=v^2\log x$. For a nonzero branch, separation givesRenaming the constant and restoring $y=v/x$ givesDividing by $v^2$ would omit the equilibrium $v=0$, so the full solution set also includes $\boxed{y\equiv 0}$. The nonzero branches are defined on connected intervals with $x>0$ and $C+x-x\log x\ne 0$. A regular solution cannot switch from a nonzero branch to zero at an ordinary point, because the right-hand side is locally Lipschitz in $v$ there. These domain restrictions are part of the full real solution.