2.10.1.4 Example 4

Solve

\[ y^{\prime }=\frac {y}{x}-\frac {2}{x}\sqrt {\sin \left ( 3\frac {y}{x}\right ) }\]

The first step is to see if we can write the above as

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }=\frac {y}{x}-\frac {2}{x}\left ( \sin \left ( 3\frac {y}{x}\right ) \right ) ^{\frac {1}{2}} \tag {2}\end{equation}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =2\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =3\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( 3\frac {y}{x}\right ) \end{align*}

Hence the solution is

\begin{equation} y=ux \tag {A}\end{equation}

Where \(u\) is the solution to

\begin{equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) ^{\frac {1}{2}} \tag {3}\end{equation}

Therefore \(f\left ( u\right ) =\sin \left ( 3u\right ) \) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-\frac {2}{x^{2}}\sin \left ( 3u\right ) ^{\frac {1}{2}}\]

This is separable.

\begin{align*} \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-\frac {2}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-2\int \frac {1}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =\frac {2}{x}+c_{1}\end{align*}

Leaving the integral as is, since it is too complicated to solve, then using \(y=ux\) where \(u\) is the solution of the above.