Question

Solve the following differential equation:

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

• A. $x^2+y^2=c(x^2-2y^2{)}^2$
• B. $x^2-y^2=c(x^2+y^2{)}^2$
• C. $y^2+x^2=c(y^2-2x^2{)}^2$
• D. $x^2-y^2=c(y^2+x^2{)}^4$

Answer 1

The correct option is D $x^2-y^2=c(y^2+x^2{)}^4$

Given differential eqn can be written as
$\frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}$
which is a homogeneous differential eqn
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, the eqn becomes
$v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$
$x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v$
$\frac{v^3-3v}{1-v^4}dv=\frac{dx}{x}$
Integrating both sides, we get
$\int \frac{v^3}{1-v^4}dv-3\int \frac{v}{1-v^4}dv=\int \frac{dx}{x}$
Put $1-v^4=t$ in first integral
$\Rightarrow -4v^{3}dv=dt$
Put $v^2=u$ in second integral
$2vdv=du$
$\frac{-1}{4}\int \frac{1}{t}dt-\frac{3}{2}\int \frac{1}{1-u^2}du=\int \frac{dx}{x}$
$\frac{-1}{4}\log t-\frac{3}{4}\log |\frac{1+u}{1-u}|=\log |x|+\log C$
$\frac{-1}{4}\log |1-v^4|-\frac{3}{4}\log |\frac{1+v^2}{1-v^2}|=\log C|x|$
$\frac{-1}{4}\log |(1-v^4)\frac{{(1+v^2)}^3}{{(1-v^2)}^3}|=\log C|x|$
$\Rightarrow \frac{{(1+v^2)}^4}{{(1-v^2)}^2}=\frac{1}{C^4{x}^4}$
$\Rightarrow C^4{x}^4{(1+v^2)}^4={(1-v^2)}^2$
$\Rightarrow c{x}^4{(1+\frac{y^2}{x^2})}^4={(1-\frac{y^2}{x^2})}^2$
$\Rightarrow (x^2-y^2{)}^2=c(x^2+y^2{)}^4$