Question

The solution of the differential equation xy2 dy - (x3 +y3) dx = 0 is

• A. $y^{3} = 3x^{3} + c$
• B. $y^{3} = 3x^{3}\log(cx)$
• C. $y^{3} = 3x^{3} + \log(cx)$
• D. $y^{3} + 3x^{3} = \log(cx)$

Answer 1

Answer: (B)

$y^{3} = 3x^{3}\log(cx)$

Answer 2

Given differential equation can be rewritten as

$\frac{dy}{dx} = \frac{x^{3} + y^{3}}{xy^{2}}$

Put $y = \nu x \Rightarrow \frac{dy}{dx}\nu + x\frac{d\nu}{dx}$

$\therefore x\frac{d\nu}{dx} = \frac{1}{\nu^{2}} = \frac{x^{3} + \nu^{3}x^{3}}{x^{3}\nu^{2}} = \frac{1 + \nu^{3}}{\nu^{2}}$

$\Rightarrow$ $x\frac{d\nu}{dx} = \frac{1}{\nu^{2}} \Rightarrow \int_{}^{}{\nu^{2}d\nu} = \int_{}^{}\frac{dx}{x} \Rightarrow \frac{\nu^{3}}{3} = \log x + \log c$

$\Rightarrow$ $\frac{1}{3}\left( \frac{y}{x} \right)^{3} = \log x + \log c \Rightarrow y^{3} = 3x^{3}\log cx$