Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 14

An integrating factor isMultiplying the linear ordinary differential equation by it givesThe right-hand side integrates to $(a^2+x^2{)}^{3/2}/3$, soThis holds on every interval where the coefficient is defined. If $a=0$, intervals cannot cross $x=0$; equivalently the family is $y=x^2/3+C/x$ on either half-line, with an independently chosen constant.

With $v=x+y$, the chain rule givesFor $v\ne 2$, this separable differential equation givesRestoring $v$ and absorbing constants, the nonconstant branches satisfyThe constant branch lost in division is $v=2$, giving the additional solution $\boxed{y=2-x}$. All branches must avoid $v=4/3$, where the original differential equation is undefined.

Yes: $y=2-x$. Its derivative is $-1$, while substitution into the right-hand side gives $-2/(6-4)=-1$. This is the equilibrium $v=2$ of the transformed autonomous differential equation; it must be restored separately after dividing by $v-2$ in the separated solution.

It is the only straight-line solution. Substituting $y=mx+d$ into the original equation shows that $m\ne -1$ would require $m=-1/3$ from the $x$ coefficients, but the constant coefficients would then give $4/3=0$. For $m=-1$, the constant equation gives $d=2$.

Let $F(p)=2p/(1-p^2)$, so that $y=xF(p)$. Differentiation givesMove the first term to the left and useSince $1+p^2>0$ for real $p$ and $p\ne \pm 1$, cancellation givesThis is the slope parameter for a homogeneous first-order equation reduction.

For $p\ne 0,\pm 1$, separate variables and use partial fraction decomposition:Integration yieldsOn any connected solution interval, absorb the fixed signs into a real constant to writeThe two real signs of the square root are chosen consistently on a branch where $C/(x+C)>0$. The constant solutions of the reduced equation are $p=0,1,-1$. The original equation excludes the last two; $p=0$ is retained by $C=0$ for $x\ne 0$.

From $p^2=C/(x+C)$, we have $1-p^2=x/(x+C)$. Substitution into the original relation gives $y=2(x+C)p$, soThese are real parabola branches, with either sign of $C$ allowed when the branch is real. For $C\ne 0$, differentiating gives $p=2C/y$ and verifies both the relation and the slope. Use graph intervals with $y\ne 0$ and $x\ne 0$, so the slope is finite and $p\ne \pm 1$. The separate constant-slope case $p=0$ gives $y=0$, also represented by the degenerate choice $C=0$. Squaring has not imposed a sign on $y$, so both real graph branches must be included.