Question

Solution of the differential equation $(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy$ is:
(where $C$ is integration constant)

• A. $(x^2+y^2)=|(x^2-y^2)|C$
• B. $(x+y{)}^2=|(x^2-y^2)|C$
• C. $(x^2+y^2{)}^3=(x^2-y^2{)}^{2}C$
• D. $(x^2+y^2{)}^2=|(x^2-y^2)C|$

Answer 1

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

The given equation can be written as:
$\frac{dy}{dx}=\frac{1-3{\left(\frac{y}{x}\right)}^2}{{\left(\frac{y}{x}\right)}^3-3\left(\frac{y}{x}\right)}$

Put $\frac{y}{x}=v$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$


$\Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}$
$\therefore \int \frac{v^3-3v}{1-v^4}dv=\int \frac{dx}{x}$
$\Rightarrow \left(\frac{-1}{4}\right)\int \frac{-4v^3}{1-v^4}dv-3\int \frac{v}{1-v^4}dv=\ln |x|+\ln |c|$
$\Rightarrow -\frac{1}{4}\ln |1-v^4|-\frac{3}{2}\int \frac{2v}{1-v^4}dv=\ln |x|+\ln |c|$
Put $v^2=t$
$\Rightarrow -\frac{1}{4}\ln |1-v^4|-\frac{3}{2}\int \frac{dt}{(1-t^2)}=\ln |x|+\ln |c|$

$\Rightarrow -\frac{1}{4}\ln |1-v^4|-\frac{3}{2}\times \frac{1}{2}\ln \left|\frac{1+t}{1-t}\right|=\ln |x|+\ln |c|$
$\Rightarrow -\frac{1}{4}\ln |x^4-y^4|+\ln |x|-\frac{3}{4}\ln \left|\frac{x^2+y^2}{x^2-y^2}\right|=\ln |x|+\ln |c|$
$\therefore \ln {\left|\frac{x^2+y^2}{x^2-y^2}\right|}^{\frac{3}{4}}+\ln {|x^4-y^4|}^{\frac{1}{4}}=\ln |c|$
$\Rightarrow (x^2+y^2)(|x^2-y^2|{)}^{\frac{-1}{2}}=|c|$
$\Rightarrow (x^2+y^2{)}^2=|(x^2-y^2)C|$