Past exam of the mathematical methods I course of the University of Oxford / 2021 / 11 / a / Solution

An integrating factor directly solves a linear first-order equation; a general second-order equation needs an additional reduction. There are several sufficient cases, with all divisions restricted to intervals where their denominators are nonzero.

If $A=0$ and $B\ne 0$, divide by $B$ to obtain $y^{\prime }+(C/B)y=D/B$. The integrating factor is $\mu =\exp (\int C/Bdx)$ andIf $C=0$ and $A\ne 0$, put $u=y^{\prime }$. Then $u^{\prime }+(B/A)u=D/A$ is first order, with integrating factor $\nu =\exp (\int B/Adx)$; solve for $u$ and integrate once more for $y$.

A further reducible class has $C=B^{\prime }-A^{\prime \prime }$, for thenOne integration produces a first-order equation for $y$, whose integrating factor after division by $A$ is $\exp (\int (B-A^{\prime })/Adx)$.

Multiplication can always put a regular second-order equation into self-adjoint form by choosing $w=\exp (\int (B-A^{\prime })/Adx)$, sinceHowever, the remaining $wCy$ term generally prevents one integration from solving it. The limitations of integrating factors for second-order equations explain why this formal transformation alone is not a general solution method. These sufficient classes do not purport to classify every specially solvable coefficient choice.