Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 5 / a

The characteristic equation of the constant-coefficient linear ordinary differential equation iswhose roots of a polynomial are $r=-1\pm i/2$. Taking real and imaginary parts of the corresponding exponential functions gives the real general solution

For a particular solution, try $y_p=a\sin x+b\cos x$. Substitution givesMatching coefficients requires $a-8b=1$ and $8a+b=8$, so $a=1$ and $b=0$. Adding the homogeneous solution from part (i), the general solution is

Put $a=\pi /2$. Substituting $x=-a$ and $x=a$ into part (ii), using $\cos (\pi /4)=\sin (\pi /4)=1/\sqrt{2}$, givesThus $A-B=\sqrt{2}(1+e^{-a})$ and $A+B=\sqrt{2}(1-e^a)$. Solving these two linear equations givesConsequently the solution of the boundary value problem isThe two independent boundary conditions determine $A,B$ uniquely.