Method of reduction of order

The method of reduction of order finds a second solution of a second-order linear homogeneous ODE when you already know one solution. Given $y_1$, look for $y_2$ in the form $y_2(x)=g(x)y_1(x)$, where $g(x)$ is to be determined.

Substituting into the ODE reduces it to a first-order equation for $g^{\prime }$, which can be solved by separation or Integrating factor.

For the standard-form ODE $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with known solution ${\varphi }_1(x)$:

$g(x)=\int \frac{e^{-\int p(x)dx}}{{\varphi }_1^2(x)}dx$

Then ${\varphi }_2(x)=g(x){\varphi }_1(x)$ is a second linearly independent solution.

Derivation

Let ${\varphi }_2=g{\varphi }_1$. Compute:

${\varphi }_2^{\prime }=g^{\prime }{\varphi }_1+g{\varphi }_1^{\prime }$

${\varphi }_2^{\prime \prime }=g^{\prime \prime }{\varphi }_1+2g^{\prime }{\varphi }_1^{\prime }+g{\varphi }_1^{\prime \prime }$

Substitute into $y^{\prime \prime }+p{y}^{\prime }+qy=0$:

$g^{\prime \prime }{\varphi }_1+2g^{\prime }{\varphi }_1^{\prime }+g{\varphi }_1^{\prime \prime }+p(g^{\prime }{\varphi }_1+g{\varphi }_1^{\prime })+qg{\varphi }_1=0$

Group:

$g^{\prime \prime }{\varphi }_1+g^{\prime }(2{\varphi }_1^{\prime }+p{\varphi }_1)+g({\varphi }_1^{\prime \prime }+p{\varphi }_1^{\prime }+q{\varphi }_1)=0$

The last bracket is zero because ${\varphi }_1$ is a solution. So:

$g^{\prime \prime }{\varphi }_1+g^{\prime }(2{\varphi }_1^{\prime }+p{\varphi }_1)=0$

Let $v=g^{\prime }$. Then $v^{\prime }=g^{\prime \prime }$, and:

${\varphi }_1{v}^{\prime }+(2{\varphi }_1^{\prime }+p{\varphi }_1)v=0$

This is a first-order separable equation for $v$. Solve:

$\frac{v^{\prime }}{v}=-\frac{2{\varphi }_1^{\prime }}{{\varphi }_1}-p$

$\ln |v|=-2\ln |{\varphi }_1|-\int pdx$

$v=\frac{e^{-\int pdx}}{{\varphi }_1^2}$

Since $g^{\prime }=v$:

$g(x)=\int \frac{e^{-\int pdx}}{{\varphi }_1^2(x)}dx$

Worked example

Find a second solution to $x^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-\frac{1}{4})y=0$ on $x>0$, given $y_1(x)=x^{-1/2}\sin x$.

Standardize: divide by $x^2$:

$y^{\prime \prime }+\frac{1}{x}{y}^{\prime }+\left(1-\frac{1}{4x^2}\right)y=0$

So $p(x)=1/x$.

Set up: $y_2=v(x)y_1$.

Substituting and simplifying (the algebra is tedious, but the structure follows the derivation above), we get:

$v^{\prime \prime }+\left(\frac{2y_1^{\prime }}{y_1}+\frac{1}{x}\right)v^{\prime }=0$

Compute $y_1^{\prime }/y_1$. With $y_1=x^{-1/2}\sin x$, derivatives give $y_1^{\prime }=-\frac{1}{2}{x}^{-3/2}\sin x+x^{-1/2}\cos x$. So:

$\frac{y_1^{\prime }}{y_1}=\frac{-\frac{1}{2}\sin x+x\cos x}{x\sin x}=\frac{\cos x}{\sin x}-\frac{1}{2x}=\cot x-\frac{1}{2x}$

So $\frac{2y_1^{\prime }}{y_1}+\frac{1}{x}=2\cot x-\frac{1}{x}+\frac{1}{x}=2\cot x$.

Reduce again: let $w=v^{\prime }$. Then $w^{\prime }=-2\cot x\cdot w$. Separate:

$\frac{w^{\prime }}{w}=-2\cot x\Rightarrow \ln |w|=-2\ln |\sin x|+C$

So $w=\frac{C_1}{{\sin }^{2}x}$, hence $v^{\prime }=w$, and integrating:

$v(x)=-C_1\cot x+C_2$

Build the solution:

$y_2=v{y}_1=(-C_1\cot x+C_2)\cdot x^{-1/2}\sin x=-C_1{x}^{-1/2}\cos x+C_2{x}^{-1/2}\sin x$

The $C_2$ term is a multiple of $y_1$ (so absorbs into $c_2$ in the general solution). The new linearly independent solution is

$y_2(x)=\frac{\cos x}{\sqrt{x}}$

The general solution: $y(x)=c_1\frac{\sin x}{\sqrt{x}}+c_2\frac{\cos x}{\sqrt{x}}$.

When to use it

Reduction of order works for second-order linear homogeneous ODEs when you have one solution and need a second. Common scenarios:

• A textbook gives you one solution and asks for the general solution.
• You guessed one solution by inspection and need the second systematically.
• The ODE has variable coefficients (so Characteristic equation doesn’t apply).

For constant-coefficient ODEs, you don’t need this method — the characteristic equation gives both solutions directly. Reduction of order shines when no closed form exists for one solution and you can only find one by inspection or special functions.

The technique generalizes: in higher-order ODEs, given $k$ solutions, you can reduce the order to find more.