Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 13

For the first-order linear ordinary differential equation, an integrating factor isMultiplication converts the left side to a derivative:Integrating gives $(1+x^2{)}^{2}y=\arctan x+C$. The initial value fixes $C=1$, so

On a branch with $y\ne 0$, the given derivative is nonzero wherever the original equation is defined, so use inverse-function linearization of a first-order equation and regard $x$ as a function of $y$:The integrating factor $1/y$ givesThus the nonzero solutions are given implicitly byEach branch must also retain $x+y+2\ne 0$; at those excluded points the original right-hand side is undefined and the inverse graph cannot represent a regular $y(x)$. Differentiating the implicit equation gives $y^{\prime }=1/(\ln |y|+C+1)$, equal to the required quotient wherever these restrictions hold.

The additional solution $y=0$, on any interval avoiding $x=-2$, was lost when dividing by $y$ and must be included. It is not defined through the point $(x,y)=(-2,0)$ in the original equation.

For $y\ne 0$, the Bernoulli differential equation becomes linear under $u=1/y$. Multiplying the equation by $-y^{-2}$ givesOn an interval with $x\ne 0$, the integrating factor is $x^{-2}$, henceTherefore the complete set of solution branches isThe original equation requires $x\ne 0$; each nonzero branch additionally requires $C+\sin x\ne 0$. The zero solution is restored explicitly because the reciprocal substitution excludes it.