Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 2 / 7B (Differential Equations)

Substitute $y_2=v{y}_1$ and use $y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1=0$ to cancel the terms proportional to $v$. The remaining ordinary differential equation isOn an interval where $y_1\ne 0$, divide through and put $w=v^{\prime }$:Thus reduction of order givesA nonzero multiplicative constant gives an independent solution; adding a constant inside the integral adds a multiple of $y_1$. The Wronskian is proportional to $e^{-\int p}$ and is nonzero on the interval. The undivided equation remains meaningful at a zero of $y_1$, but this integral representation then needs continuation rather than direct division.

Inspection gives $y_1=x$ in the degree-one Legendre differential equation. After division by $1-x^2$, $p=-2x/(1-x^2)$ and $e^{-\int p}=1/(1-x^2)$. On either side of zero, reduction of order therefore givesA second solution is $y_2=x\mathrm{arctanh}x-1$. Although $v$ is singular at zero, the product $v{y}_1$ has a smooth continuation throughout $(-1,1)$. This illustrates reduction of order across a zero of the known solution. Since $y_1(0)=0$, $y_1^{\prime }(0)=1$, $y_2(0)=-1$ and $y_2^{\prime }(0)=0$, the required coefficients are $1,-1$. HenceThe ordinary differential equation has regular coefficients near zero, so the resulting initial-value solution is unique and the removable singularity in the intermediate representation causes no ambiguity.