The general solution of (x+1)dydx+1=2e−y is

ey(x+1)=x+C

e−y=2x+C

ey(x+1)=2x+C

ey(x+1)=C

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[c] (x+1)dydx+1=2e−y ⇒eydydx+eyx+1=2x+1 Put ey=u⇒eydydx=dudx∴dudx+ux+1=2x+1I.F. =e∫1x+1dx=elog(x+1)=x+1 ∴ Solution is u(x+1)=∫2(x+1)(x+1)dx+C ⇒ey(x+1)=2x+C

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Question Text	The general solution of (x+1)dydx+1=2e−y is
Answer Type	Text solution:1
Upvotes	151

The general solution of (x+1)dydx+1=2e−y is

Text solutionVerified

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