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

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(B) Given, $(x+y+1) \frac{dy}{dx} = 1$ $\Rightarrow \frac{dx}{dy} = x+y+1$ $\Rightarrow \frac{dx}{dy} - x = y+1$, which is a linear differential equat

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$x = -(y+2) + ce^y$

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(B) Given, $(x+y+1) \frac{dy}{dx} = 1$ $\Rightarrow \frac{dx}{dy} = x+y+1$ $\Rightarrow \frac{dx}{dy} - x = y+1$, which is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$. Here, $P = -1$ and $Q = y+1$. The Integrating Factor $(IF)$ is $e^{\int P dy} = e^{\int -1 dy} = e^{-y}$. The solution is given by $x \cdot (IF) = \int Q \cdot (IF) dy + c$. $x e^{-y} = \int (y+1) e^{-y} dy + c$ Using integration by parts for $\int y e^{-y} dy$: $x e^{-y} = [y(-e^{-y}) - \int 1 \cdot (-e^{-y}) dy] + \int e^{-y} dy + c$ $x e^{-y} = -y e^{-y} - e^{-y} - e^{-y} + c$ $x e^{-y} = -(y+2) e^{-y} + c$ Multiplying both sides by $e^y$, we get: $x = -(y+2) + ce^y$.

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

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