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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.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For real parameters, work on $x>0$ so arbitrary real powers are unambiguous. Taking derivatives givesThe residual after substitution isWhen $\gamma \ne 0$, $x^{\gamma }$ ranges over $(0,\infty )$. Evaluating at the zeros of the cosine first forces the sine coefficient to vanish; then the cosine coefficient vanishes identically. If $\gamma =0$, the nonzero constant $\cos 1$ cannot satisfy the equation. Thus the necessary and sufficient conditions areUnder these conditions the normalized coefficient is $p=(1-\gamma )/x$. The Abel identity gives $W=C{x}^{\gamma -1}$. Choose $C=\gamma$ and apply reduction of order on an interval where $\cos (x^{\gamma })\ne 0$:The substitution $u=x^{\gamma }$ evaluates the integral. After continuation across the cosine zeros,This is a power substitution for an oscillatory second-order equation: in the variable $u=x^{\gamma }$, the equation reduces to the harmonic oscillator equation.

Here the indicial equation has the repeated root $0$. The first Frobenius method solution is $y_1=e^{-x}$. Since $p=1+1/x$, the Abel identity gives $W=C{e}^{-x}/x$. Reduction of order with $C=1$ produceswhere the additive constant is absorbed into $y_1$. In particular the logarithmic coefficient is nonzero: an additional independent real analytic solution does not exist.

The general local form isThis is the repeated-root Logarithmic Frobenius solution. On the negative real side one may use $\log |x|$, or use a fixed logarithm branch for a complex solution.

Differentiating the Wronskian givesThus $W^{\prime }+pW=0$, the Abel identity.

Also,If $y_2(x_0)=0$, integration gives the reduction of order formula

For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution iswhich is a nonzero multiple of $x^{-1}$. HenceThe conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so

After division by $x$, the coefficient of $y^{\prime }$ is $P(x)=1/x-1$. The Abel identity for the Wronskian givesso

For $\lambda =1$, direct substitution confirms that $y_1=1-x$. Reduction of order gives a second solution proportional toTo extract the requested coefficients, putFor the differential operatorone findsThe analytic correction $h={\sum }_{n\ge 1}{b}_n{x}^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients giveHence