Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 9

Use separation of variables on intervals avoiding zeros of $\cos x$:Taking the tangent gives the real general solutionEach solution is restricted to an interval where both the original equation and this expression are finite. There is no real constant solution because $1+y^2$ never vanishes for real $y$.

For $x\ne 0$, divide by $x$ and use the integrating factor $\mu (x)=e^x|x{|}^{\alpha }$, separately on each side of zero. The equation becomes $(\mu y{)}^{\prime }=|x{|}^{\alpha }/x$. ThusFor $x>0$ the absolute values may be omitted. The original equation is singular as a first-order evolution equation at zero, so these formulas describe regular intervals with $x\ne 0$; extension through zero needs a separate regularity condition.

Rearranging gives $y^{\prime }=2x(y-1{)}^2$. First retain the equilibrium $y=1$. For every other solution, separation of variables gives $\int (y-1{)}^{-2}dy=\int 2xdx$, so $-1/(y-1)=x^2+C$. HenceA nonconstant solution is defined on any interval where its denominator is nonzero. Dividing by $(y-1{)}^2$ without retaining the equilibrium would lose one solution.