The solution of frac{dy}{dx} + 1 = e^{x+y} is | vedclass

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Question

(A) Given differential equation is $\frac{dy}{dx} + 1 = e^{x+y}$.Let $x + y = z$.Differentiating both sides with respect to $x$,we get $1 + \frac{dy}{

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$e^{-(x+y)} + x + c = 0$

Vedclass · Answer

(A) Given differential equation is $\frac{dy}{dx} + 1 = e^{x+y}$.Let $x + y = z$.Differentiating both sides with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$.Substituting this into the original equation,we have $\frac{dz}{dx} = e^z$.Separating the variables,we get $e^{-z} dz = dx$.Integrating both sides,$\int e^{-z} dz = \int dx$.This gives $-e^{-z} = x + c$.Substituting $z = x + y$ back,we get $-e^{-(x+y)} = x + c$.Rearranging the terms,we get $x + e^{-(x+y)} + c = 0$.

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

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