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

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.