# Question #b0376

Jan 20, 2018

$\frac{x y}{{x}^{2} - {y}^{2}} ^ 2 = {c}_{1}$

#### Explanation:

Let $y = v x$, where v is a function of x, which means $\frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$. The given DE , thus transforms to $v + x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{{v}^{3} + 3 v}{1 + 3 {v}^{2}}$

$x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{{v}^{3} + 3 v}{1 + 3 {v}^{2}} - v = \frac{2 v - 2 {v}^{3}}{1 + 3 {v}^{2}}$

$\frac{1 + 3 {v}^{2}}{v \left(1 - {v}^{2}\right)} \mathrm{dv} = 2 \left(\frac{1}{x}\right) \mathrm{dx}$. Partial fractions on the left side would be

$\left(\frac{1}{v} + \frac{4 v}{1 - {v}^{2}}\right) \mathrm{dv} = 2 \left(\frac{1}{x}\right) \mathrm{dx}$. Now integrate on both sides to get

$\ln v - 2 \ln \left(1 - {v}^{2}\right) = 2 \ln x + C$. On simplification it becomes,

$\ln \left(\frac{v}{1 - {v}^{2}} ^ 2\right) = \ln {x}^{2} + C$
$\frac{v}{1 - {v}^{2}} ^ 2 = {c}_{1} {x}^{2}$

$\frac{y {x}^{3}}{{x}^{2} - {y}^{2}} ^ 2 = {c}_{1} {x}^{2}$

$\frac{x y}{{x}^{2} - {y}^{2}} ^ 2 = {c}_{1}$