Question

The solution of the D.E. $(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy$, is:

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

Answer 1

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

Given differential equation
$(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy$
$\Rightarrow \frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}$ ...(1)
which is a homogeneous differential equation
So, put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, eqn (1) becomes,
$v+x\frac{dv}{dx}=\frac{x^3-3v^2{x}^3}{v^3{x}^3-3x^{3}v}$
$\Rightarrow v+x\frac{dv}{dx}==\frac{1-3v^2}{v^3-3v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}$
$\Rightarrow \frac{v^3-3v}{1-v^4}dv=\frac{dx}{x}$
Integrating both sides,
$\int \frac{v^3-v}{1-v^4}dv-\int \frac{2v}{1-v^4}dv=\int \frac{dx}{x}$
$\Rightarrow \int \frac{v(v^2-1)}{(1-v^2)(1+v^2)}dv-\int \frac{2v}{1-(v^2{)}^2}dv=\int \frac{dx}{x}$
$\int \frac{-v}{(1+v^2)}dv-\int \frac{2v}{1-(v^2{)}^2}dv=\int \frac{dx}{x}$
Put $v^2=t$ ,
$\Rightarrow 2vdv=dt$
$\int \frac{-dt}{2(1+t)}-\int \frac{dt}{1-t^2}=\log x+\log c$
$-\frac{1}{2}\log |1+t|-\frac{1}{2}\log |\frac{1+t}{1-t}=\log x+\log c$
$\Rightarrow -\log |1+t|+\frac{1}{2}\log |1-t|=\log x+\log c$
$\Rightarrow -\log |1+v^2|+\frac{1}{2}\log |1-v^2|=\log x+\log c$
$\Rightarrow -\log |x^2+y^2|+\log x^2+\frac{1}{2}\log |y^2-x^2|-\frac{1}{2}\log x^2=\log x+\log c$
$\Rightarrow \log \frac{\sqrt{y^2-x^2}}{x^2+y^2}=\log c$
$\Rightarrow \frac{\sqrt{y^2-x^2}}{x^2+y^2}=c$
$\Rightarrow y^2-x^2=C(x^2+y^2{)}^2$