Find the general solution of the differential equation: $(x^3 + y^3) dy = x^2y dx$

$\frac{x^3}{3y^3} = \log y + c$

The correct answer is Option (2) → $\frac{x^3}{3y^3} = \log y + c$ ##

Given differential equation is

$(x^3 + y^3) dy = x^2y dx$

$∴\frac{dx}{dy} = \frac{x^3 + y^3}{x^2y} \quad \dots(i)$

Put $x = vy \Rightarrow \frac{dx}{dy} = v + y \frac{dv}{dy}$

From Eq. (i), we have

$v + y \frac{dv}{dy} = \frac{(vy)^3 + y^3}{(vy)^2y}$

$v + y \frac{dv}{dy} = \frac{v^3y^3 + y^3}{v^2y^3}$

$v + y \frac{dv}{dy} = \frac{v^3 + 1}{v^2}$

$y \frac{dv}{dy} = \frac{v^3 + 1}{v^2} - v$

$y \frac{dv}{dy} = \frac{1}{v^2}$

$v^2 dv = \frac{dy}{y} \quad (\text{variable separation method})$

Integrating both sides, we get

$\int v^2 dv = \int \frac{dy}{y}$

$\frac{v^3}{3} = \log y + c$

Putting $v = \frac{x}{y}$, we get

$\frac{x^3}{3y^3} = \log y + c$