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
Expert's answer
2021-08-23T04:48:30-0400

Using homogeneous method

let y=vxdydx=(x3+y3)x2ydydx=(x3+(vx)3)x2(vx)dydx=(x3+v3x3)x3vdydx=x3(1+v3)x3vdydx=(1+v3)vTherefordydx=1+vdvdx1+vdvdx=(1+v3)vvdvdx=(1+v3)v(1)dydx=(1v+v2)1dydx=(1/v2+v1/v)dv(v2+v1v)=dx(v33)+(lnv)(v22)=xbutv=yx(yx)33+(ln(yx))(yx)22=x+clet \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


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