Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 7

The characteristic equation of the associated homogeneous linear ordinary differential equation is $m^2+5m+6=(m+2)(m+3)=0$, so the complementary function is $C_1{e}^{-2x}+C_2{e}^{-3x}$. For a particular solution, try $y_p=ax+b$. Substitution gives $6ax+5a+6b=6x$, so $a=1$, $b=-5/6$. The unrestricted solution is thereforeIts derivative at zero is $-2C_1-3C_2+1$. The single prescribed condition imposes $2C_1+3C_2=1$ and leaves one arbitrary constant. Thus the most general permitted solution isThere is no condition fixing $y(0)$, so the remaining arbitrary constant is essential.

Write $y(\theta )=Y(x)$ with $x=\cos \theta$. By the chain rule,The terms involving $Y_x$ cancel in the original ordinary differential equation, leavingOn each interval where $\sin \theta \ne 0$, the transformed linear ordinary differential equation is $Y_{xx}+2Y=0$. Its characteristic equation has roots $\pm i\sqrt{2}$, yieldingDirect substitution shows these expressions also extend smoothly through $\theta =k\pi$ and satisfy the original equation there. The two constants describe the general solution on any interval where the coordinate substitution is invertible.