Prove that ${x^2} - {y^2} = c{\left( {{x^2} + {y^2}} \right)^2}$ is the general solution of differential equation, $\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy$ where $c$ is a parameter.

$\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{{x^3} - 3x{y^2}}}{{{y^3} - 3{x^2}y}}$

…..(1)
This is a homogeneous equation. To simplify it, we need to make the substitution as:$y = vx$
$\Rightarrow$ $\frac{d}{{dx}}(y) = \frac{d}{{dx}}(vx)$

$\Rightarrow$ $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

Substituting the values of $y$ and $\frac{{dv}}{{dx}}$
$v + x\frac{{dv}}{{dx}} = \frac{{{x^3} - 3x{{(vx)}^2}}}{{{{(vx)}^3} - 3{x^2}(vx)}}$
$\Rightarrow$ $v + x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2}}}{{{v^3} - 3v}}$

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

$\Rightarrow$ $x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2} - v\left( {{v^3} - 3v} \right)}}{{{v^3} - 3v}}$

$\Rightarrow$ $x\frac{{dv}}{{dx}} = \frac{{1 - {v^4}}}{{{v^3} - 3v}}$

$\Rightarrow$ $\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)dv = \frac{{dx}}{x}$
Integrating both sides,

we get:
$\int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = \log x + \log {C^\prime }$

…..(2)

Now, $\int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = \int {\frac{{{v^3}dv}}{{1 - {v^4}}}} - 3\int {\frac{{vdv}}{{1 - {v^4}}}}$
$\Rightarrow$ $\int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = {I_1} - 3{I_2}$, where ${I_1} = \int {\frac{{{v^3}dv}}{{1 - {v^4}}}}$ and ${I_2} = \int {\frac{{vdv}}{{1 - {v^4}}}}$……(3)

Let $1 - {v^4} = t$
$\therefore \frac{d}{{dv}}\left( {1 - {v^4}} \right) = \frac{{dt}}{{dv}}$
$\Rightarrow$ $- 4{v^3} = \frac{{dt}}{{dv}}$
$\Rightarrow$ ${v^3}dv = - \frac{{dt}}{4}$

Now, ${I_1} = \int {\frac{{ - dt}}{{4t}}} = - \frac{1}{4}\log t = - \frac{1}{4}\log \left( {1 - {v^4}} \right)$

And, ${I_2} = \int {\frac{{vdv}}{{1 - {v^4}}}} = \int {\frac{{vdv}}{{1 - {{\left( {{v^2}} \right)}^2}}}}$

Let ${v^2} = p$
$\therefore \frac{d}{{dv}}\left( {{v^2}} \right) = \frac{{dp}}{{dv}}$

$\Rightarrow$ $2v = \frac{{dp}}{{dv}}$

$\Rightarrow$ $vdv = \frac{{dp}}{2}$

$\Rightarrow$ ${I_2} = \frac{1}{2}\int {\frac{{dp}}{{1 - {p^2}}}} = \frac{1}{{2 \times 2}}\log \left| {\frac{{1 + p}}{{1 - p}}} \right| = \frac{1}{4}\log \mid \frac{{1 + {v^2}}}{{1 - {v^2}}}$

Substituting the values of ${{\rm{I}}_1}$ and ${{\rm{I}}_2}$ in equation $(3)$,

we get:
$\int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = - \frac{1}{4}\log \left( {1 - {v^4}} \right) - \frac{3}{4}\log \left| {\frac{{1 - {v^2}}}{{1 + {v^2}}}} \right|$

Therefore, equation (2) becomes:
$\frac{1}{4}\log \left( {1 - {v^4}} \right) - \frac{3}{4}\log \left| {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right| = \log x + \log {{\rm{C}}^\prime }$

$\Rightarrow$ $- \frac{1}{4}\log \left[ {\left( {1 - {v^4}} \right){{\left( {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right)}^3}} \right] = \log {{\rm{C}}^\prime }x$

$\Rightarrow$ $\frac{{{{\left( {1 + {v^2}} \right)}^4}}}{{{{\left( {1 - {v^2}} \right)}^2}}} = {\left( {{{\rm{C}}^\prime }x} \right)^{ - 4}}$
$\Rightarrow$ $\frac{{{{\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right)}^ + }}}{{{{\left( {1 - \frac{{{y^2}}}{{{x^2}}}} \right)}^2}}} = \frac{1}{{{{\rm{C}}^{\prime 4}}{x^4}}}$

$\Rightarrow$ $\frac{{{{\left( {{x^2} + {y^2}} \right)}^ + }}}{{{x^4}{{\left( {{x^2} - {y^2}} \right)}^2}}} = \frac{1}{{{{\rm{C}}^{\prime 4}}{x^4}}}$

$\Rightarrow$ ${\left( {{x^2} - {y^2}} \right)^2} = C{{\rm{'}}^4}{\left( {{x^2} + {y^2}} \right)^4}$

$\Rightarrow$ $\left( {{x^2} - {y^2}} \right) = C{'^2}{\left( {{x^2} + {y^2}} \right)^2}$

$\Rightarrow$ ${x^2} - {y^2} = C{\left( {{x^2} + {y^2}} \right)^2}$, where $C = {C^{\prime 2}}$
Hence, the given result is proved.