Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 13

For a first-order linear ordinary differential equation $y^{\prime }+p(x)y=q(x)$, an integrating factor is $\exp (\int p(x)dx)$. Here $p(x)=\cot x$, soThe sine is positive on the specified interval, so this integrating factor never vanishes there. Any nonzero constant multiple is equally suitable.

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.

Insert the exponential function trial $y=e^{rx}$ into the homogeneous linear differential equation. Its characteristic equation is $r^2+\sqrt{3}r+3=0$, with polynomial rootsTaking real combinations of the two complex exponential solutions produces the general solution

Use the method of undetermined coefficients with a particular solution $K{e}^{-\sqrt{3}x}$. The differential operator multiplies this exponential function by $3-3+3=3$. Matching the forcing coefficient gives $3K=1$, hence

A linear polynomial trial $y=ux+v$ has $y^{\prime }=u$ and $y^{\prime \prime }=0$. Substituting gives $3ux+(\sqrt{3}u+3v)=x$, so $u=1/3$ and $v=-\sqrt{3}/9=-1/(3\sqrt{3})$. Thus a particular solution is

The superposition principle gives the general solution as $y_c+2y_{p,2}+y_{p,1}$:The first boundary condition gives $A=2/(3\sqrt{3})-1/3$. At $x=\pi$, the cosine vanishes and the sine equals $-1$. Comparing with the second boundary condition then givesThe unique solution satisfying both boundary conditions is thereforeUniqueness follows because a homogeneous solution zero at both endpoints would have $A=0$ from the first and $B=0$ from the second.