The solution of the equation frac{dy}{dx} = f | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = \frac{1}{x+y+1}$.Let $v = x+y+1$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx}

Vedclass · Accepted Answer

$y = \log(x+y+2) + c$,where $c$ is the constant of integration

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(C) Given the differential equation $\frac{dy}{dx} = \frac{1}{x+y+1}$.Let $v = x+y+1$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting these into the original equation: $\frac{dv}{dx} - 1 = \frac{1}{v}$.Rearranging the terms: $\frac{dv}{dx} = 1 + \frac{1}{v} = \frac{v+1}{v}$.Separating the variables: $\frac{v}{v+1} dv = dx$.Integrating both sides: $\int \frac{v+1-1}{v+1} dv = \int dx$.This simplifies to $\int (1 - \frac{1}{v+1}) dv = \int dx$.Integrating gives $v - \log|v+1| = x + c$.Substituting $v = x+y+1$ back into the equation: $(x+y+1) - \log|x+y+1+1| = x + c$.This simplifies to $y+1 - \log|x+y+2| = c$,or $y = \log|x+y+2| + C'$,where $C' = c-1$ is a constant.Thus,the correct option is $C$.

The solution of the equation $\frac{dy}{dx} = \frac{1}{x+y+1}$ is

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