Question

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

• A. $x^{3}-y^{3}=3kxy +C$
• B. $x^{3}+y^{3}=3kxy +C$
• C. $x^{2}-y^{2}=2kxy +C$
• D. $x^{2}+y^{2}=2kxy +C$
• E. $x^{3}-y^{2}=3kxy +C$

Answer 1

Answer: (B)

Given differential equation is
$\left(k x-y^{2}\right) d y=\left(x^{2}-k y\right) d x$
$\Rightarrow k x\, d y-y^{2} d y=x^{2} d x-k y\, d x$
$\Rightarrow k(x \,d y+y d x)=x^{2} \,d x+y^{2} \,d y$
$\Rightarrow k[d(x y)]=x^{2} \,d x+y^{2}\, d y$
On integrating both sides, we get
$k(x y)=\frac{x^{3}}{3}+\frac{y^{3}}{3}-\frac{C}{3}$
$ \Rightarrow x^{3}+y^{3}=3\, k x y+C$