Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 16

Set $v=y^{\prime }$ to reduce the second-order linear differential equation to $x{v}^{\prime }+v=0$. Equivalently, $(xv{)}^{\prime }=0$, so $v=C_1/x$ on any interval excluding zero. The derivative condition at $x=1$ gives $C_1=2$.

Integrating on the positive interval containing the initial point yields $y=2\ln x+C_2$. The value $y(1)=-1$ then fixes $C_2=-1$, givingIts derivative is $2/x$, and direct substitution verifies the equation. The selected solution diverges as $x\downarrow 0$ and cannot extend through zero as a finite-valued function.

Construct the scalar potential $F(x,y)=y{e}^{x^2}-x^2$. Its differential iswhich proves exactness directly. Multiplying the differential equation by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $y{e}^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. ThereforeThe factor $e^{x^2}$ is also an integrating factor for the associated first-order linear equation.

The complementary function is $C\cos 2x+D\sin 2x$. A particular solution is $(\sin x+\cos x)/3$, since applying $d^2/d{x}^2+4$ to either sine or cosine multiplies it by $3$. ThusAt zero, $C+1/3=3$, giving $C=8/3$. At $x=\pi /4$, the cosine term vanishes and the particular term is $\sqrt{2}/3$, so $D=-\sqrt{2}/3$. ConsequentlyThe two boundary conditions determine the two constants uniquely because their evaluation matrix on $\cos 2x,\sin 2x$ is the identity.