Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 6A (Differential Equations)

Substitute the power series and compare coefficients. The constant coefficient gives $-a_1-a_0=0$. For $k\ge 1$, the coefficient of $x^k$ givesThus $a_1=-a_0$, while the $k=1$ equation places no restriction on $a_2$. For $k\ge 2$,The power series therefore sums toRenaming the two arbitrary constants gives $\boxed{y(x)=C(1-x)+D{e}^{-x}}$.

Both functions solve the equation by direct substitution. Their Wronskian, in the order $1-x,e^{-x}$, is $x{e}^{-x}$, which is nonzero for $x\ne 0$. Hence they give the full two-dimensional solution space on either interval away from the singular point. Both are entire, and the displayed formula also gives every twice continuously differentiable solution through zero: matching $y(0)$ and $y^{\prime \prime }(0)$ fixes the same $C,D$ on the two sides.

For two solutions $y_1,y_2$, define their Wronskian byDifferentiation and substitution of the second-order linear differential equation giveThis proves the Abel identity and hence

On an interval where the known solution $y_1$ does not vanish,Integrating gives the reduction of order formulaChoosing the normalization $C=1$ makes the Wronskian nonzero, so this solution is linearly independent of $y_1$. To verify it really solves the equation, write $y_2=y_{1}h$. Its equation reduces tothe integrand gives $h^{\prime }=\exp (-\int p)/y_1^2$, which satisfies this equation on differentiation. Adding an integration constant only adds a multiple of $y_1$. The formula is local where $y_1\ne 0$; at a zero of $y_1$ with regular coefficients, the actual solution can be continued by initial-value existence and uniqueness, even if its integral representation has an apparent singularity.

Multiply the equation by $x$ to obtain $x{y}^{\prime \prime }-P(x)y^{\prime }-Q(x)y=0$. The leading terms after a Frobenius method trial $y=x^r$ arethe term involving $Q_0$ first appears at order $x^r$. The indicial equation is therefore

To obtain the first correction to a regular singular solution by reduction of order, start from the assumed solution $y_1=1+\beta x+O(x^2)$ and use the Abel identity:We work initially on $x>0$ near zero; since $P_0$ is a positive integer, the resulting expansion is an ordinary power series. Also $y_1^{-2}=1-2\beta x+O(x^2)$, so the reduction of order integral has integrandChoose its integration constant to be zero, eliminating an added multiple of $y_1$. Integrating and multiplying by $y_1$ givesWith $C=P_0+1$ the requested normalized answer isIts nonzero Wronskian makes it independent of $y_1$. The existence of the asserted $y_1$ also forces $P_0\beta +Q_0=0$; there is no need to introduce $Q_0$ into the normalized second coefficient.