2.10.1.6 Example 6

Solve

\[ y^{\prime }=\frac {y}{x}+x\sin \left ( \frac {y}{x}\right ) \]

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}+x\sin \left ( \frac {y}{x}\right ) \tag {2}\end{equation}

Comparing (2) to (1) shows that

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

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

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

This is separable.

\begin{align*} \frac {1}{\sin u}du & =dx\\ \int \frac {1}{\sin u}du & =\int dx\\ \ln \sin \frac {u}{2}-\ln \cos \frac {u}{2} & =x+c_{1}\\ \ln \tan \frac {u}{2} & =x+c_{1}\\ \tan \frac {u}{2} & =c_{2}e^{x}\\ \frac {u}{2} & =\arctan \left ( c_{2}e^{x}\right ) \\ u & =2\arctan \left ( c_{2}e^{x}\right ) \end{align*}

Hence (A) becomes

\[ y=2x\arctan \left ( c_{2}e^{x}\right ) \]