4.10.1.5 Example 5

Solve

\[ y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x=0 \]

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{align} y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x & =0\nonumber \\ y^{\prime }x & =y-2x^{3}\tan \left ( \frac {y}{x}\right ) \nonumber \\ y^{\prime } & =\frac {y}{x}-2x^{2}\tan \left ( \frac {y}{x}\right ) \tag {2}\end{align}

Comparing (2) to (1) shows that

\begin{align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-2x^{2}\\ b & =1\\ f\left ( b\frac {y}{x}\right ) & =\tan \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 \(f\left ( u\right ) =\tan u\) and \(\left ( 3\right ) \) becomes

\[ u^{\prime }=-2x\tan u \]

This is separable.

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

Hence (A) becomes

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