Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 13

The homogeneous linear differential equation is $y^{\prime \prime }+y=0$. Its characteristic equation is ${\lambda }^2+1=0$, with roots $\pm i$. The general real complementary function is thereforeThe two trigonometric solutions are linearly independent, supplying the two arbitrary constants required for this second-order linear ordinary differential equation.

Use the method of undetermined coefficients, choosing a particular solution $y_p=A+B\cos 2x$. Substitution givesMatching the constant and cosine coefficients to $\frac{1}{2}-\frac{1}{2}\cos 2x$ gives $A=1/2$, $B=1/6$. ThusNeither forcing term is resonant with the homogeneous angular frequency one.

Combine the complementary function and particular solution:The boundary condition at zero gives $C_1+2/3=0$, while that at $\pi /2$ gives $C_2+1/3=0$. HenceThe two boundary conditions determine both constants uniquely; substitution at each endpoint gives zero.

First retain the equilibrium solution $\boxed{y\equiv 0}$. Any nonzero real solution has constant sign on its interval of existence, since the polynomial right-hand side is locally Lipschitz and Picard-Lindelöf theorem prevents crossing the zero solution.

For a nonzero solution, the Bernoulli differential equation substitution $u=y^{-4}$ givesThis linear ordinary differential equation has solution $u+2=C{e}^{-8x}$. Therefore every nonzero real solution isEither sign may be chosen and remains fixed. The restriction ensures $y^{-4}>0$; nonpositive $C$ gives no nonzero real solution. These are maximal real intervals, and their restrictions are also solutions. As the right endpoint is approached, $|y|$ diverges in finite-time blowup. The substitution would have lost $y=0$ if it had not been checked separately.