Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 12

Use variation of parameters, writing $y_p=t_1{u}_1+t_2{u}_2$ and imposing $t_1^{\prime }{u}_1+t_2^{\prime }{u}_2=0$. Differentiation then gives $y_p^{\prime }=t_1{u}_1^{\prime }+t_2{u}_2^{\prime }$. Since the $u_j$ solve the homogeneous equation,Thus solve the two-by-two linear systemIts determinant is the nonzero Wronskian $W$. Cramer's rule gives $t_1^{\prime }=-u_{2}R/W$ and $t_2^{\prime }=u_{1}R/W$, and integration gives the stated antiderivatives. Their integration constants only add homogeneous solutions. The construction applies on a regular interval where the forcing and coefficients are defined and $W\ne 0$.

Differentiate the Wronskian and cancel the mixed first-derivative terms:This is the constant-coefficient form of Abel's identity. Hence $W(x)=C{e}^{-ax}$, or $W(x)=W(x_0)e^{-a(x-x_0)}$. Since the exponential function is never zero, $W$ is either identically zero or nowhere zero on a connected interval of the homogeneous equation.

Linear dependence means there are constants $c_1,c_2$, not both zero, with $c_1{u}_1+c_2{u}_2=0$. Since $u_1$ is nowhere zero, $c_2\ne 0$, so $u_2=C{u}_1$ and $u_2^{\prime }=C{u}_1^{\prime }$. Substitution gives $W=u_{1}C{u}_1^{\prime }-C{u}_1{u}_1^{\prime }=0$.

The nonvanishing assumption is more than is needed for this direction. It also gives the converse directly: if $W=0$, then $(u_2/u_1{)}^{\prime }=W/u_1^2=0$, so the ratio is constant on a connected interval. For arbitrary differentiable functions with zeros, a vanishing Wronskian alone needs more care; here the homogeneous linear equation also supplies an appropriate uniqueness theorem.

The forcing $\sec (2x)$ is undefined at $x=\pm \pi /4$, both inside the printed interval. Therefore there is no classical solution on the whole stated interval. Solve separately on the three connected components obtained by removing those two points.

Choose $u_1=\cos (2x)$ and $u_2=\sin (2x)$, whose Wronskian is two. Variation of parameters gives $t_1^{\prime }=-(1/2)\tan (2x)$ and $t_2^{\prime }=1/2$, so a particular solution is $\frac{1}{4}\cos (2x)\log |\cos (2x)|+\frac{x}{2}\sin (2x)$. The full answer on each regular component isThe constants $A,B$ can differ between $(-\pi /2,-\pi /4)$, $(-\pi /4,\pi /4)$ and $(\pi /4,\pi /2)$. Differentiation gives $y^{\prime \prime }+4y=1/\cos (2x)$ wherever the forcing is defined. Although the particular solution has finite one-sided values at the singularities, its derivative has a logarithmic divergence, and the equation itself is undefined there. It cannot be promoted to one classical solution across the poles.