Question

The general solution of the differential equation $\left(x^3-3x{y}^2\right)dx=\left(y^3-3x^{2}y\right)dy$ is

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

Answer 1

Answer: (A)

The given differential equation is
$\left(x^3-3x{y}^2\right)dx=\left(y^3-3x^{2}y\right)dy$
$\frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}....$(i)
So, put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes
$\Rightarrow v+x\frac{dv}{dx}=\frac{x^3-3x(vx{)}^2}{(vx{)}^3-3x^2(vx)}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v$
$\Rightarrow x\frac{dv}{dx}=\frac{1-3v^2-v^4+3v^2}{v^3-3v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}$
$\Rightarrow \left(\frac{v^3-3v}{1-v^4}\right)dv=\frac{dx}{x}$
On integrating both sides, we get
$\int \left(\frac{v^3-3v}{1-v^4}\right)dv=\int \frac{dx}{x}$
$\Rightarrow \int \left(\frac{v^3-3v}{1=v^4}\right)dv=\log x+\log C_1....(ii)$
Now, $\int \left(\frac{v^3-3v}{1-v^4}\right)dv=\int \frac{v^3}{1-v^4}dv-3\int \frac{v}{1-v^4}dv$
$\Rightarrow \int \left(\frac{v^3-3v}{1-v^4}\right)dv=I_1-3I_2.....$(iii)
where, $I_1=\int \frac{v^3}{1-v^4}dv$ and $I_2=\int \frac{v}{1-v^4}dv$
Put $1-v^4=t\Rightarrow -4v^3=\frac{dt}{dv}\Rightarrow v^{3}dv=-\frac{dt}{4}$
$\therefore l_1=\int \frac{-dt}{4t}=-\frac{1}{4}\log t=-\frac{1}{4}\log \left(1-v^4\right)$ and
and $I_2=\int \frac{vdv}{1-v^4}=\int \frac{vdv}{1-{\left(v^2\right)}^2}$
Put $v^2=z\Rightarrow vdv=\frac{dz}{2}$
$\Rightarrow l_2=\frac{1}{2}\int \frac{dz}{1-z^2}=\frac{1}{2\times 2}\log \left|\frac{1+z}{1-z}\right|$
$=\frac{1}{4}\log \left|\frac{1+v^2}{1-v^2}\right|\left(\because z=v^2\right)$
$\left(\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \mid \frac{a+x}{a-x}\right)$
On substituting the values of $I_1$ and $I_2$ in Eq. (iii), 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, Eq. (ii) becomes $-\frac{1}{4}\log \left(1-v^4\right)-\frac{3}{4}\log \left|\frac{1+v^2}{1-v^2}\right|=\log x+\log C_1$
$\Rightarrow -\frac{1}{4}\log \left[\left(1-v^4\right){\left(\frac{1+v^2}{1-v^2}\right)}^3\right]=\log C_{1}x$
$\Rightarrow -\frac{1}{4}\log \left[\left(1-v^2\right)\left(1+v^2\right)\times \frac{{\left(1+v^2\right)}^3}{{\left(1-v^2\right)}^3}\right]=\log C_{1}x$ $\Rightarrow \log {\left[\frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}\right]}^{-\frac{1}{4}}=\log C_{1}x$
$\Rightarrow {\left[\frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}\right]}^{-\frac{1}{4}}=C_{1}x$
$\Rightarrow \frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}={\left(C_{1}x\right)}^{-4}\Rightarrow \frac{{\left(1+\frac{y^2}{x^2}\right)}^4}{{\left(1-\frac{y^2}{x^2}\right)}^2}=\frac{1}{C^4{x}^4}$
$\Rightarrow \frac{{\left(x^2+y^2\right)}^4}{x^4{\left(x^2-y^2\right)}^2}=\frac{1}{C_1^4{x}^4}(\because v=y/x)$
$\Rightarrow {\left(x^2-y^2\right)}^2=C_1^4{\left(x^2+y^2\right)}^4$
$x^2-y^2=C{\left(x^2+y^2\right)}^2($ where, $C=C_1^2)$