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Question

The solution of the equation (2x+y+1)dx+(4x+2y1)dy=0 is

A
log(2x+y1)=C+x+y
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B
log(4x+2y1)=C+2x+y
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C
log(2x+y+1)+x+2y=C
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D
log(2x+y1)+x+2y=C
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Solution

The correct option is D log(2x+y1)+x+2y=C
dydx=2x+y+14x+2y1

Let t=2x+y

dtdx=2+dydx

substituting in above equations;

2dtdx=t+12t1

dtdx=3t32t1

(2t1)dt=3(t1)dx

2dt+dtt1=3dx

Integrate on both sides;

2t+log(t1)=3x+C

2(2x+y)+log(2x+y1)=3x+C

log(2x+y1)+x+2y=C

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