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# Answer to Question #225920 in Chemical Engineering for Lokika

Question #225920

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

1
2021-08-23T04:48:30-0400

Using homogeneous method

"let \\space y=vx\\\\\n\n\u21d2\\frac{dy}{dx}=\\frac{(x\u00b3+y\u00b3)}{x\u00b2y}\\\\\n\n\\frac{dy}{dx}=\\frac{(x\u00b3+ (vx)\u00b3)}{x\u00b2(vx)}\\\\\n\n\u21d2 \\frac{dy}{dx}=\\frac{(x\u00b3+v\u00b3x\u00b3)}{x\u00b3v}\\\\\n\n\\frac{dy}{dx}=\\frac{x\u00b3(1+v\u00b3)}{x\u00b3v}\\\\\n\n\\frac{dy}{dx}=\\frac{(1+v\u00b3)}{v}\\\\\n\nThere for \\frac{dy}{dx}=\\frac{1+vdv}{dx}\\\\\n\n\u21d2 \\frac{1+vdv}{dx}=\\frac{(1+v\u00b3)}{v}\n\n\u21d2\\frac{vdv}{dx}=\\frac{(1+v\u00b3)}{v(-1)}\\\\\n\n\\frac{dy}{dx}=(\\frac{1}{v+v\u00b2})-1\\\\\n\n\\frac{dy}{dx}=(1\/v\u00b2+v-1\/v)\\\\\n\n\u21d2\\frac{dv}{(v\u00b2+v\u2227-1-v)}=dx\\\\\n\n\u21d2 (\\frac{v\u00b3}{3})+(\\ln v)-(\\frac{v\u00b2}{2})=x\\\\\n\nbut v=\\frac{y}{x}\\\\\n\n\u21d2 \\frac{(\\frac{y}{x})\u00b3}{3} + (\\ln(\\frac{y}{x})) - \\frac{(\\frac{y}{x})\u00b2}{2}=x+c"

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