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Monday, December 28, 2015

Homogenous Equation

Problem:

(y2x2)dx+xydy=0

Solution:

(y2x2)dx+xydy=0(y2x2)dx=xydydydx=x2y2xy=xyyx

Now let z=yx,dzdx=ddx(y)1x+yddx1x=dydx1xyx2, which implies dydx=xdzdx+z. Putting these back, we have

dydx=xyyxxdzdx+z=1zzxdzdx=1z2zzdz12z2=dxx14log(12z2)=log(x)+C1log(12z2)=4log(x)+C212z2=C3x412(yx)2=C3x4x22y2=C3x2y2=12(x2Cx2)

For a quick check, we have:

y2=12(x2Cx2)2ydydx=x+Cx3=x+Cx2x=x+x22y2x=2x2y2xdydx=xyyx

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