Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 9 / c

For $b\ne -1$ and initially $y\ne 0$, use the Bernoulli power substitution $v=y^{b+1}$. The equation becomesWrite $\kappa =a(b+1)$. A first-order linear differential equation then givesThe second line is the resonant case of a Bernoulli equation with exponential forcing. Recover $y$ using real branches for which the original power $y^{-b}$ is defined; for example, unrestricted noninteger powers are unambiguous on $y>0$.

The exceptional exponent $b=-1$ makes the original equation linear: $y^{\prime }=(a+e^{-2x})y$. Its solutions areDomain restrictions matter when the solution reaches zero. If $b<0$ and the real power at zero is defined, $y\equiv 0$ is also a solution. For $-1<b<0$, waiting-time solutions of a sublinear Bernoulli equation may stay zero until an arbitrary $x_0$ and then become positive, withThe joining derivative is zero because $1/(b+1)>1$. For $b=0$, the linear formula can pass through zero normally; for $b>0$, the original negative power excludes zero. These qualifications prevent dividing out a valid zero solution or inventing one where the equation is undefined.