Question

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