Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 6A / Solution

With the indicated order of the two solutions, define the Wronskian byWhere $y_1\ne 0$, this gives the first-order linear differential equationIntegrating produces the reduction of order formulaThis formula is local on an interval where the known solution does not vanish; an arbitrary nonzero scaling of the linearly independent solution changes the normalization of $W$.

Differentiate the Wronskian and use the differential equation for each solution:Thus $\boxed{W^{\prime }+pW=0}$, proving the Abel identity, and $W=C\exp (-\int pdx)$. For continuous coefficients on a nonsingular interval, linearly independent solutions have $C\ne 0$.

For the particular equation, write $s=x-1$. The proposed solution is $y_1=s^2$, with $y_1^{\prime }=2s$ and $y_1^{\prime \prime }=2$, so substitution gives $2s^2+2s^2-4s^2=0$. Away from the singular point $x=1$, division by $s^2$ gives $p=1/s$. The Abel identity therefore giveswith a separate constant on each interval lying on one side of the singular point. The reduction of order integral is $\int W/y_1^{2}dx=C\int s^{-5}ds=-C/(4s^4)$, soChoosing $C_2=0$, $C=-4$ givesThe nonzero Wronskian proves linear independence. Direct substitution also verifies the second solution, since $s^2(6s^{-4})+s(-2s^{-3})-4s^{-2}=0$. These solutions form a basis on either nonsingular interval; the second solution is not defined at $x=1$.