Past exam of the mathematical methods I course of the University of Oxford / 2022 / 11

Separation of variables gives $\ln |y|=\sqrt{1+x^2}+C$ for a nonzero solution. Including the zero solution in the arbitrary real amplitude givesDifferentiating the exponent verifies the equation. This form keeps the zero solution that a division by $y$ could otherwise conceal.

On the component $x>0$ containing the boundary point, divide by $2x$ and use integrating factor $x^{-1/2}$:Integration gives $y=x^3/5+C\sqrt{x}$, and $y(1)=1$ fixes $C=4/5$. ThereforeAlthough the function has a continuous limit at zero, its derivative diverges there; the solution with this boundary datum does not extend as a differentiable solution through the singular point of the equation.

The exponential substitution removing a constant first derivative, $y=e^{x}u$, reduces the equation to $u^{\prime \prime }+{\beta }^{2}u=1+\cos x$. For $\beta \ne 0$ and ${\beta }^2\ne 1$, solve the homogeneous harmonic equation and use a constant plus a cosine for the particular solution:There is a further exceptional case inside the printed condition ${\beta }^2\ne 1$: $\beta =0$ must be treated separately. It gives $u^{\prime \prime }=1+\cos x$, henceThis retains the constant forcing's resonance with the zero-frequency homogeneous equation. Substitution verifies both branches.

For ${\beta }^2=1$, the reduced equation is $u^{\prime \prime }+u=1+\cos x$. The constant forcing has particular solution one, while the resonant cosine requires $u_p=(x/2)\sin x$. ThusThe secular factor $x$ is the signature of resonance. It must replace the divergent nonresonant coefficient $1/({\beta }^2-1)$; substituting ${\beta }^2=1$ directly into that coefficient is invalid.