Reduction of order

Definition

for given $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$

If $y_1$ is a solution, let $y_2=y:=u{y}_1$ Then, $y^{\prime }=u^{\prime }{y}_1+u{y}_1^{\prime },y^{\prime \prime }=u^{\prime \prime }{y}_1+2u^{\prime }{y}_1^{\prime }+u{y}_1^{\prime \prime }$

Substitute $y^{\prime },y^{\prime \prime }$ in the original expression $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$

$(u^{\prime \prime }{y}_1+2u^{\prime }{y}_1^{\prime }+u{y}_1^{\prime \prime })+p(x)(u^{\prime }{y}_1+u{y}_1^{\prime })+q(x)y=0$ $\Rightarrow u^{\prime \prime }{y}_1+u^{\prime }(2y_1^{\prime }+p{y}_1)+u\underset{=0(\because y_1=\text{sol})}{\underbrace{(y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1)}}=0$ $\Rightarrow u^{\prime \prime }{y}_1+u^{\prime }(2y_1^{\prime }+p{y}_1)=0$ $\Rightarrow u^{\prime \prime }+u^{\prime }(\frac{2y_1^{\prime }+p{y}_1}{y_1})=0$ $\Rightarrow U’+U(\frac{2y_1^{\prime }+p{y}_1}{y_1}+p)=0,\text{where}U(x):=u^{\prime }(x)$ $\Rightarrow \frac{dU}{U}-(\frac{2y_1^{\prime }+p{y}_1}{y_1}+p)dx$ $\Rightarrow \ln |U|=-\int (\frac{2y_1^{\prime }+p{y}_1}{y_1}+p)dx=-2\ln |y_1|-\int pdx$ $\Rightarrow U=\frac{1}{y_1^2}{e}^{-\int pdx}$ $\therefore y_2=u{y}_1=y_1\int Udx$