Answer to Question #225920 in Chemical Engineering for Lokika

Question

The integral factor of (x²y)dx- (x³ + y³)dy = 0 is

A) -y^4

B) -1/y^4

C) y^4 D) 1/y^4

Answer 1

Using homogeneous method

$let \space y=vx\\ ⇒\frac{dy}{dx}=\frac{(x³+y³)}{x²y}\\ \frac{dy}{dx}=\frac{(x³+ (vx)³)}{x²(vx)}\\ ⇒ \frac{dy}{dx}=\frac{(x³+v³x³)}{x³v}\\ \frac{dy}{dx}=\frac{x³(1+v³)}{x³v}\\ \frac{dy}{dx}=\frac{(1+v³)}{v}\\ There for \frac{dy}{dx}=\frac{1+vdv}{dx}\\ ⇒ \frac{1+vdv}{dx}=\frac{(1+v³)}{v} ⇒\frac{vdv}{dx}=\frac{(1+v³)}{v(-1)}\\ \frac{dy}{dx}=(\frac{1}{v+v²})-1\\ \frac{dy}{dx}=(1/v²+v-1/v)\\ ⇒\frac{dv}{(v²+v∧-1-v)}=dx\\ ⇒ (\frac{v³}{3})+(\ln v)-(\frac{v²}{2})=x\\ but v=\frac{y}{x}\\ ⇒ \frac{(\frac{y}{x})³}{3} + (\ln(\frac{y}{x})) - \frac{(\frac{y}{x})²}{2}=x+c$