3.9.8.5.2 Example 8. \(4x^{2}\sin \left ( x\right ) y^{\prime \prime }+\left ( -4x^{2}\cos x-4x\sin x\right ) y^{\prime }+\left ( 2x\cos x+3\sin x\right ) y=0\)

Writing the ode in normal form gives

\begin{align*} y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y & =0\\ p & =\frac {-4x^{2}\cos x-4x\sin x}{4x^{2}\sin \left ( x\right ) }\qquad x\neq 0,\pi ,2\pi ,\cdots \\ q & =\frac {2x\cos x+3\sin x}{4x^{2}\sin \left ( x\right ) }\\ r & =0 \end{align*}

Applying transformation on the dependent variable second method \(y=v\left ( x\right ) x^{n}\) results in

\begin{align} x^{n}v^{\prime \prime }+\left ( 2nx^{n-1}+px^{n}\right ) v^{\prime }+\left ( n\left ( n-1\right ) x^{n-2}+px^{n-1}n+qx^{n}\right ) v & =0\nonumber \\ v^{\prime \prime }+\frac {\left ( 2nx^{n-1}+px^{n}\right ) }{x^{n}}v^{\prime }+\left ( \frac {n\left ( n-1\right ) x^{n-2}+px^{n-1}n+qx^{n}}{x^{n}}\right ) v & =0\nonumber \\ v^{\prime \prime }+\left ( 2nx^{-1}+p\right ) v^{\prime }+\left ( n\left ( n-1\right ) x^{-2}+px^{-1}n+q\right ) v & =0\nonumber \\ v^{\prime \prime }+\left ( 2nx^{-1}+p\right ) v^{\prime }+\left ( pnx^{-1}+q+\left ( n^{2}-n\right ) x^{-2}\right ) v & =0 \tag {1}\end{align}

Assuming the coefficient of \(v\left ( x\right ) \) above is zero. This gives

\[ pnx^{-1}+q+\left ( n^{2}-n\right ) x^{-2}=0 \]

Substituting the values for \(p,q\,\)in the above gives

\[ \left ( \frac {-4x^{2}\cos x-4x\sin x}{4x^{2}\sin \left ( x\right ) }\right ) nx^{-1}+\frac {2x\cos x+3\sin x}{4x^{2}\sin \left ( x\right ) }+\left ( n^{2}-n\right ) x^{-2}=0 \]

Solving for \(n\) shows that \(n=\frac {1}{2}\). Hence (1) now reduces to

\begin{align*} v^{\prime \prime }+\left ( x^{-1}+p\right ) v^{\prime } & =0\\ v^{\prime \prime }+\left ( \frac {1}{x}+\frac {-4x^{2}\cos x-4x\sin x}{4x^{2}\sin \left ( x\right ) }\right ) v^{\prime } & =0\\ v^{\prime \prime }+\left ( \frac {4x\sin x-4x^{2}\cos x-4x\sin x}{4x^{2}\sin x}\right ) v^{\prime } & =0\\ v^{\prime \prime }+\left ( \frac {-4x^{2}\cos x}{4x^{2}\sin x}\right ) v^{\prime } & =0\\ v^{\prime \prime }-\frac {\cos x}{\sin x}v^{\prime } & =0 \end{align*}

Let \(v^{\prime }=u,\) the above becomes

\[ u^{\prime }-\frac {\cos x}{\sin x}u=0 \]

Which is linear first order ode. It has the solution \(u=c_{1}\sin \left ( x\right ) \). Hence

\[ v^{\prime }=c_{1}\sin \left ( x\right ) \]

Integrating gives

\[ v=-c_{1}\cos \left ( x\right ) +c_{2}\]

Therefore

\begin{align*} y & =v\sqrt {x}\\ & =\left ( -c_{1}\cos \left ( x\right ) +c_{2}\right ) \sqrt {x}\end{align*}

This can also be written as

\[ y=\left ( c_{3}\cos \left ( x\right ) +c_{2}\right ) \sqrt {x}\]