For the given differential equation find the general solution.

$(x + y) \frac{d y}{d x} = 1$

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Solution

$(x + y) \frac{d y}{d x} = 1 \Rightarrow \frac{d x}{d y} = (x + y) \Rightarrow \frac{d x}{d y} - x = y$
On comparing with the form $\frac{d x}{d y} + P x = Q,$ we get
P=-1, Q=y
$∴ I F = e^{\int P d y} = e^{- \int 1 d y} \Rightarrow I F = e^{- y}$ ....(i)
The general solution of the given differential equations is given by the relation,
$x . I F = \int Q \times I F d y + C \Rightarrow x . e^{- y} = \int e^{- y} . y d y + C \Rightarrow x e^{- y} = y \int e^{- y} d y - \int [\frac{d}{d y} y \int e^{- y} d y] d y + C \Rightarrow x e^{- y} = - y (e^{- y}) + \int e^{- y} d y + C \Rightarrow x e^{- y} = - y e^{- y} - e^{- y} + C \Rightarrow x = - y - 1 + e^{y} C \Rightarrow x + y + 1 = C e^{y} .$

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