91Ó°ÊÓ

The given equation is,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{3\mathrm{xy}}{\mathbf{2}{\mathbf{x}}^2\mathbf{-}{\mathbf{y}}^2}$

Substitute the $\mathbf{y}\mathbf{=}\mathbf{vx}\text{ a²Ô»å }\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}$in the above equation,

$\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{3}\mathbf{x}\left(\mathrm{vx}\right)}{\mathbf{2}{\mathbf{x}}^2\mathbf{-}{\left(\mathrm{vx}\right)}^2}$

Simplify the above equation,

$\begin{array}{c}\mathbf{v}\mathbf{+}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{3v}{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}\\ \mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{3v}{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}\mathbf{-}\mathbf{v}\\ \mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{3}\mathbf{v}\mathbf{-}\mathbf{2}\mathbf{v}\mathbf{+}{\mathbf{v}}^3}{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}\\ \mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}\end{array}$

Cross multiplication on both sides in the above equation,

$\begin{array}{c}\mathbf{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}\\ \frac{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}\mathbf{dv}\mathbf{=}\frac{\mathrm{dx}}{x}\end{array}$

Integration,

$∫\frac{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}\mathbf{dv}=∫\frac{\mathrm{dx}}{x}$

Solve the above equation,

$\begin{array}{c}∫\frac{\mathbf{2}\mathbf{-}{\mathbf{v}}^2}{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}\mathbf{dv}\mathbf{=}∫\frac{\mathrm{dx}}{x}\\ ∫\frac{2}{\mathbf{v}\mathbf{+}{\mathbf{v}}^3}\mathbf{dv}\mathbf{-}\frac{1}{2}∫\frac{2v}{\mathbf{1}\mathbf{+}{\mathbf{v}}^2}\mathbf{=}∫\frac{\mathrm{dx}}{x}\\ \mathbf{2}\mathbf{ln}\left|v\right|\mathbf{-}\mathbf{ln}\left|{\mathbf{v}}^2\mathbf{+}\mathbf{1}\right|\mathbf{-}\frac{1}{2}\mathbf{ln}\left|{\mathbf{v}}^2\mathbf{+}\mathbf{1}\right|\mathbf{=}\mathbf{ln}\text{ }\mathbf{x}\mathbf{+}\mathbf{ln}\text{ }\mathbf{c}\\ \mathbf{2}\mathbf{ln}\left|v\right|\mathbf{-}\frac{3}{2}\mathbf{ln}\left|{\mathbf{v}}^2\mathbf{+}\mathbf{1}\right|\mathbf{=}\mathbf{ln}\text{ }\mathbf{x}\mathbf{+}\mathbf{ln}\text{ }\mathbf{c}\end{array}$

Substitute the $\mathbf{v}\mathbf{=}\frac{y}{x}$ in the above equation,

$\begin{array}{c}\mathbf{2}\mathbf{ln}\left|\frac{y}{x}\right|\mathbf{-}\frac{3}{2}\mathbf{ln}\left|{\frac{y}{{\mathbf{x}}^2}}^2\mathbf{+}\mathbf{1}\right|\mathbf{=}\mathbf{ln}\text{ }\mathbf{x}\mathbf{+}\mathbf{ln}\text{ }\mathbf{c}\\ \mathbf{ln}\left({\left(\frac{y}{x}\right)}^2\right)\mathbf{-}\mathbf{ln}\left({\left(\frac{{\mathbf{y}}^2}{{\mathbf{x}}^2}\mathbf{+}\mathbf{1}\right)}^{\frac{3}{2}}\right)\mathbf{-}\mathbf{ln}\left(x\right)\mathbf{=}\mathbf{ln}\left(c\right)\\ \mathbf{ln}\left(\frac{{\left(\frac{y}{x}\right)}^2}{{\left(\frac{{\mathbf{y}}^2}{{\mathbf{x}}^2}\mathbf{+}\mathbf{1}\right)}^{\frac{3}{2}}}\right)\mathbf{-}\mathbf{ln}\left(x\right)\mathbf{=}\mathbf{ln}\left(c\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}\mathbf{ln}\left(\frac{\frac{{\mathbf{y}}^2\mathbf{x}}{{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^{\frac{3}{2}}}}{x}\right)\mathbf{=}\mathbf{ln}\left(c\right)\\ \mathbf{ln}\left(\frac{{\mathbf{y}}^2}{{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^{\frac{3}{2}}}\right)\mathbf{=}\mathbf{ln}\left(c\right)\\ \frac{{\mathbf{y}}^2}{{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^{\frac{3}{2}}}\mathbf{=}\mathbf{c}\\ {\mathbf{y}}^2\mathbf{=}\mathbf{c}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^{\frac{3}{2}}\end{array}$

Taking square on both sides,

$\begin{array}{c}{\left({\mathbf{y}}^2\right)}^2\mathbf{=}{\left(\mathbf{c}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^{{}^{\frac{3}{2}}}\right)}^2\\ {\mathbf{y}}^4\mathbf{=}\mathbf{C}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^3\end{array}$

Where, C = c²

Hence, solution of the given differential equation,

${\mathbf{y}}^4\mathbf{=}\mathbf{C}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^3$

Let C = 1 then ${\mathbf{y}}^4\mathbf{=}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^3$.

Let C = 2 then ${\mathbf{y}}^4\mathbf{=}\mathbf{2}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^3$.

Let C = 3 then ${\mathbf{y}}^4\mathbf{=}\mathbf{3}{\left({\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^2\right)}^3$.

Therefore, for different values of C, the sketch of the solution curves is as shown below: