Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 15

The characteristic equation has the repeated polynomial root $m=-2$, because both $e^{-2x}$ and $x{e}^{-2x}$ occur as independent solutions. Hence its characteristic polynomial isComparing constant terms gives

The characteristic equation factors asThe distinct characteristic roots therefore give the general solution

The characteristic roots areTaking real linear combinations of the two complex exponential solutions yieldsThe coefficient $\beta$ is three here, as confirmed in the PDF; the OCR replaces it by an undefined symbol.

The associated homogeneous linear differential equation has characteristic roots $2\pm i$, so its complementary function is $e^{2x}(A\cos x+B\sin x)$. For a particular integral, the method of undetermined coefficients suggests $y_p=a{x}^2+bx+c$. Substitution and comparison of coefficients givesso $y_p=25x^2+40x+22$. The general solution is thereforeThe first boundary condition gives $A=-21$. At $x=\pi /2$, the polynomial part already equals the assigned boundary value and the cosine vanishes, so $e^{\pi }B=0$. ThusThis is the unique solution satisfying the two boundary conditions; here the phrase “particular solution” means that selected member of the full family, rather than only the polynomial particular integral.