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
$\left(kx-y^2\right)dy=\left(x^2-ky\right)dx$
$\Rightarrow kxdy-y^{2}dy=x^{2}dx-kydx$
$\Rightarrow k(xdy+ydx)=x^{2}dx+y^{2}dy$
$\Rightarrow k[d(xy)]=x^{2}dx+y^{2}dy$
On integrating both sides, we get
$k(xy)=\frac{x^3}{3}+\frac{y^3}{3}-\frac{C}{3}$
$\Rightarrow x^3+y^3=3kxy+C$