4.10.1.6 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

$$\begin{array}{}\text{(1)}& y^{\prime }=\frac{y}{x}+g\left(x\right)f{\left(b\frac{y}{x}\right)}^{\frac{n}{m}}\end{array}$$

Hence

$$\begin{array}{}\text{(2)}& y^{\prime }=\frac{y}{x}+x\sin \left(\frac{y}{x}\right)\end{array}$$

Comparing (2) to (1) shows that

$$\begin{array}{rl}n& =1\\ m& =1\\ g\left(x\right)& =x\\ b& =1\\ f\left(b\frac{y}{x}\right)& =\sin \left(\frac{y}{x}\right)\end{array}$$

Hence the solution is

$$\begin{array}{}\text{(A)}& y=ux\end{array}$$

Where $u$ is the solution to

$$\begin{array}{}\text{(3)}& u^{\prime }=\frac{1}{x}g\left(x\right)f\left(u\right)\end{array}$$

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.

$$\begin{array}{rl}\frac{1}{\sin u}du& =dx\\ \int \frac{1}{\sin u}du& =\int dx\\ \ln \sin \frac{u}{2}-\ln \cos \frac{u}{2}& =x+c_1\\ \ln \tan \frac{u}{2}& =x+c_1\\ \tan \frac{u}{2}& =c_2{e}^x\\ \frac{u}{2}& =\arctan \left(c_2{e}^x\right)\\ u& =2\arctan \left(c_2{e}^x\right)\end{array}$$

Hence (A) becomes

$$y=2x\arctan \left(c_2{e}^x\right)$$