Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 7

On an interval avoiding $x=\pm 1$, separation of variables gives, for $y\ne -1$,Absorb the sign into the integration constant and include the equilibrium solution $y=-1$ by allowing $C=0$. The complete real family isEach solution is considered on a connected interval in $\mathbb{R}\setminus \{-1,1\}$, where the original equation is defined.

Divide by $\cos x$ on an interval where it is nonzero. For the resulting first-order linear differential equation, use the integrating factor $M=\sec x+\tan x$, since $M^{\prime }=\sec xM$. The right-hand side after division is also $M^{\prime }$, soThe identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, givingAn integrating factor may have either fixed sign; if obtained by integrating $\sec x$, the real logarithm is $\ln |\sec x+\tan x|$.

The Bernoulli differential equation becomes linear after $z=1/y$: multiplying by $-y^{-2}$ gives $z^{\prime }-2z=-x^2$. A quadratic particular solution is $z_p=x^2/2+x/2+1/4$, while the homogeneous solution is $C{e}^{2x}$. The initial condition gives $z(0)=1$ and $C=3/4$. HenceThe denominator is positive for all real $x$, since $2x^2+2x+1=2(x+1/2{)}^2+1/2$. Thus this initial value problem has the displayed solution on the entire real line. The separate solution $y=0$ does not satisfy its initial condition.