Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 20 / a

For the solution through $x=0$, work on the connected domain $x>-1$. While $y>0$, separation of variables givesThe initial condition gives $C=2$, soThe restriction matters: the unsquared equation requires $1-\frac{1}{2}\ln (1+x)\ge 0$. The solution reaches zero at $x_{\ast }=e^2-1$. Continuing the squared formula beyond that point would give a positive derivative, contradicting the original negative square-root right-hand side.

The nonnegative forward continuation is insteadThe two pieces join with $y^{\prime }=0$, and both satisfy the original equation. Because the right-hand side is nonpositive for $x>-1$, a nonnegative solution cannot leave zero in the forward direction. This is a finite-time extinction phenomenon for a separable differential equation.

This is a Bernoulli differential equation with exponent four. On the nonzero branch selected by $y(0)=1$, put $v=y^{-3}$. Multiplying the original equation by $-3y^{-4}$ givesIts integrating factor is $e^{-x}$, so $(e^{-x}v{)}^{\prime }=-3$ and $e^{-x}v=C-3x$. The initial condition sets $C=1$. ThusThis is the maximal real solution interval containing zero: the solution diverges as $x\to (1/3{)}^{-}$. The real negative branch of the same algebraic formula for $x>1/3$ is disconnected from this initial-value solution. The original equation also admits $y=0$, but it does not satisfy the prescribed initial condition.