Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 16

The separable differential equation gives $d(1-y)/(1-y)=-3dx/x$. Thus $1-y=C{x}^{-3}$ on $x>0$, and the initial value gives $C=1$. HenceThe initial-value branch cannot cross its singularity at zero.

Multiplication by $e^y$ gives $d(e^y)/dx=x/\sqrt{1+x^2}$. Integrating and imposing $y(0)=0$ yields $e^y=\sqrt{1+x^2}$. Therefore

On an interval avoiding zero put $v=y/x$. A real solution requires $|v|\le 1$, and the principal square root is $\sqrt{x^2-y^2}=|x|\sqrt{1-v^2}$. The homogeneous first-order differential equation reduces toWhere $|v|<1$, separation of variables gives $\arcsin v=\mathrm{sgn}(x)\log |x|+C$. Thus the nonconstant pieces arerestricted to intervals where the cosine of the phase is nonnegative. This sign condition is essential when substituting back into the original square root.

There are also the solutions $y=x$ and $y=-x$. They can join the nonconstant pieces with a continuous first derivative, because the transformed rate vanishes but is not Lipschitz continuous at $v=\pm 1$. For example, on the positive axis every $A>0$ givesThe sticking solutions of a bounded square-root differential equation describe the same joining mechanism on either half-axis after using the increasing variable $t=\mathrm{sgn}(x)\log |x|$.

Near the initial point, division by $\cos x$ gives $y^{\prime }+(1+\tan x)y=2\sec x+2\sin x$. An integrating factor is $e^x\sec x$, andThus $y=2\sin x+C{e}^{-x}\cos x$, and $y(0)=2$ givesAlthough division by $\cos x$ was local, direct substitution verifies this smooth solution in the original equation for every real $x$, including the zeros of $\cos x$.

Set $u=1/y$, so $u^{\prime }=-y^{\prime }/y^2$. The equation becomes $x(x^2-1)u^{\prime }+u=-x^3$. For $x>1$ its integrating factor is $\sqrt{x^2-1}/x$, givingThe initial value is $u(2)=-2$, forcing $C=0$. HenceSubstitution into the original equation verifies this answer throughout $x>0$, including $x=1$ where the divided linear equation was singular. Its maximal connected interval through the initial point is $(0,\infty )$.