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Question

The solution of the differential equation x2dy+y(x+y)dx=0 is:

A
x2y=c2(2x+y)
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B
xy2=c2(2x+y)
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C
x2y=c2(x+2y)
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D
xy2=c2(x+2y)
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Solution

The correct option is B x2y=c2(2x+y)
Given differential equation can be re-written as follows:
(xy)2dydx+xy+1=0 -------(i)
Substitute y=xt
dydx=t+dtdx
With above substitution equation (i) will become:
1t2(t+xdtdx)+1t+1=0xdtdx=(2t+t2)dtt(t+2)=dxx
12[1t1t+2]dt=dxx
12ln(tt+2)=lnx+c1
ln(tt+2)=2lnx+lnk, where k=e2c1
ln(tt+2)=ln(kx2)
tt+2=kx2
Back substituting t=yx in above equation, we will get
y(2x+y)=kx2
x2y=c2(2x+y), where k=c2

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