3.3.7.6 Example 6

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{-3yx}{3x^2+y^2}\\ y\left(0\right)& =1\end{array}$$

At $x=0,y=1$ then $f(x,y)=\frac{-3y-x}{3x^2+y^2}$ is defined. And $f_y$ is also defined at $x=0,y=1$. Hence a unique solution exist.

Let $y=ux$ or $u=\frac{y}{x}$, hence $\frac{dy}{dx}=x\frac{du}{dx}+u$ and the above ode becomes

$$\begin{array}{rl}x\frac{du}{dx}+u& =\frac{-3u{x}^2}{3x^2+u^2{x}^2}\\ x\frac{du}{dx}+u& =\frac{-3u}{3+u^2}\\ x\frac{du}{dx}& =\frac{-3u}{3+u^2}-u\\ & =\frac{-3u-u(3+u^2)}{3+u^2}\\ & =\frac{-6u-u^3}{3+u^2}\end{array}$$

This is separable.

$$\frac{3+u^2}{-6u-u^3}du=\frac{1}{x}dx$$

Integrating

$$\begin{array}{rl}\int \frac{3+u^2}{-6u-u^3}du& =\int \frac{1}{x}dx\\ -\frac{1}{2}\ln u-\frac{1}{4}\ln (u^2+6)& =\ln x+c\\ -\frac{1}{2}\ln u-\frac{1}{4}\ln (u^2+6)-\ln x& =\ln x+\ln c_1\end{array}$$

Solving for $u$ gives

$$\begin{array}{rl}{u}_1& =-\sqrt{-3-\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ u_2& =\sqrt{-3-\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ u_3& =-\sqrt{-3+\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ u_4& =\sqrt{-3+\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\end{array}$$

Hence

$$\begin{array}{rl}\frac{y_1}{x}& =-\sqrt{-3-\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ \frac{y_2}{x}& =\sqrt{-3-\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ \frac{y_3}{x}& =-\sqrt{-3+\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\\ \frac{y_4}{x}& =\sqrt{-3+\frac{1}{2}\sqrt{36+\frac{4}{x^4{c}_1^4}}}\end{array}$$

or for $x\ge 0$

$$\begin{array}{rl}{y}_1& =-\sqrt{-3x^2-\frac{1}{2}\sqrt{36x^4+\frac{4}{c_1^4}}}\\ y_2& =\sqrt{-3x^2-\frac{1}{2}\sqrt{36x^4+\frac{4}{c_1^4}}}\\ y_3& =-\sqrt{-3x^2+\frac{1}{2}\sqrt{36x^4+\frac{4}{c_1^4}}}\\ y_4& =\sqrt{-3x^3+\frac{1}{2}\sqrt{36x^4+\frac{4}{c_1^4}}}\end{array}$$

Applying IC $y\left(0\right)=1$

$$\begin{array}{rl}-1& =\sqrt{-\sqrt{\frac{1}{c_1^4}}}\\ 1& =\sqrt{-\sqrt{\frac{1}{c_1^4}}}\\ -1& =\sqrt{\sqrt{\frac{1}{c_1^4}}}\\ 1& =\sqrt{\sqrt{\frac{1}{c_1^4}}}\end{array}$$

or

$$\begin{array}{rl}-1& =\sqrt{\frac{1}{c_1^4}}\\ -1& =\sqrt{\frac{1}{c_1^4}}\\ 1& =\sqrt{\frac{1}{c_1^4}}\\ 1& =\sqrt{\frac{1}{c_1^4}}\end{array}$$

Throwing the first 2 since complex. Then $c_1=1$. Hence

$$\begin{array}{rl}y& =\sqrt{-3x^3+\frac{1}{2}\sqrt{36x^4+4}}\\ & =\sqrt{-3x^3+\sqrt{9x^4+1}}\end{array}$$