Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 8

For the Bernoulli differential equation, work on a branch where $y\ne 0$ and the required powers are defined. Multiplication by $(1-m)y^{-m}$ makes the chain rule useful:Thus $\alpha =1-m$, $h=(1-m)f$ and $k=(1-m)g$. The case $m=1$ is already linear and must be treated separately, because the proposed substitution would be constant. Solutions lost by dividing by a power of $y$, particularly $y=0$ when it solves the original equation, must also be checked separately.

Here $m=2$, so $z=1/y$ obeys $z^{\prime }-2z\tan x=-{\tan }^{2}x$. An integrating factor is ${\cos }^{2}x$, givingThe nonzero solutions are thereforeThe additional solution is $\boxed{y=0}$. Each formula is interpreted on an interval where $\tan x$ is defined and its displayed denominator is nonzero.

The given particular solution $y_p=x^2$ suggests the standard reduction of a Riccati equation. Substitution of $y=x^2+1/v$ givesOn either side of $x=0$, an integrating factor is $|x{|}^{5/2}$, so $v=(C-2\sqrt{|x|})/|x{|}^{5/2}$. Hencealong with the separately recovered particular solution $y=x^2$. The denominator must remain nonzero on the solution interval; $x=0$ is a singular point of the divided differential equation.

For the positive-$x$ branch, $y(1)=(C-1)/(C-2)$ and $y(4)=16(C-2)/(C-4)$. Nonsingular endpoint values exclude $C=2,4$, soThe particular solution $y=x^2$ gives the same value directly. These evaluations concern one solution branch, with a common integration constant, rather than independently chosen solutions at the two points.