The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the solutionA. `(y-1)(e^(x)-1) = Ce^(y)`B. ` (y-1)(e^(x)+1) = Ce^(y)`C. ` (y+1) (e^(x)-1)=Ce^(y)`D. ` (y+1)(e^(x)+1)= Ce^(y)`

The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the s

1.	The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the solutionA. `(y-1)(e^(x)-1) = Ce^(y)`B. ` (y-1)(e^(x)+1) = Ce^(y)`C. ` (y+1) (e^(x)-1)=Ce^(y)`D. ` (y+1)(e^(x)+1)= Ce^(y)`
Answer» Correct Answer - c Given differential equation can be rewritten as ` (y dy)/(y+1) = (e^(x)dx)/(e^(x)+1)` ` rArr (1- 1/(y+1))dy = (e^(x))/(e^(x)+1)dx` ` rArr y - log (y+1) = log (e^(x)+1) - log C` [ integrating both sides ] ` rArr y = log. ((e^(x)+1)(y+1))/C` , ` rArr (e^(x)+1) (y+1) = Ce^(y)`

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The differential equation ` (e^(x)+1)y dy = (y+1) e^(x) dx ` has the solutionA. `(y-1)(e^(x)-1) = Ce^(y)`B. ` (y-1)(e^(x)+1) = Ce^(y)`C. ` (y+1) (e^(x)-1)=Ce^(y)`D. ` (y+1)(e^(x)+1)= Ce^(y)`

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