Question

The solution of $(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy$ is:

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

Answer 1

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

Given $(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy$...........(1)
Dividing LHS and RHS by $x^3$ and taking $\frac{y}{x}$ equal to $v$ we have,
$\frac{dy}{dx}=v+x\frac{dv}{dx}$ .....(2)
Using (2) in (1) we have,$v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$
$\Rightarrow$$x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v$
$\Rightarrow$$x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}$
$\Rightarrow$$x\frac{(v^3-3v)dv}{v^4-1}=\frac{-dx}{x}$
Let $\frac{v^3-3v}{v^4-1}=\frac{Av+B}{v^2-1}+\frac{Cv+D}{v^2+1}$
Now solving for $A,B,C$ and $D$ we have $A=-1$, $C=2$ and $B=D=0$
Hence we have, $\frac{v^3-3v}{v^4-1}=\frac{-v}{v^2-1}+\frac{2v}{v^2+1}$
Now, $\Rightarrow$$\int \frac{(v^3-3v)dv}{v^4-1}=\int \frac{-dx}{x}$
$\Rightarrow$$\int (\frac{-vdv}{v^2-1}+\frac{2vdv}{v^2+1})=\int \frac{-dx}{x}$
$\Rightarrow$$\frac{1}{2}\int (\frac{-2vdv}{v^2-1}+\frac{4vdv}{v^2+1})=\int \frac{-dx}{x}$
$\Rightarrow$ $-ln|v^2-1|+2ln|v^2+1|=-2ln|x|+k$
Now putting the value of $v$ we have ,
$\frac{(y^2+x^2{)}^2}{x^2(y^2-x^2)}=\frac{c}{x^2}$ where $c$ is a constant.
Finally, $y^2-x^2=c(y^2+x^2{)}^2$ where $c$ is a constant.
Hence, option 'A' is correct.