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The solution of $(x+y+1) \frac{d y}{d x}=1$ is
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Verified Answer
The correct answer is:
$x=-(y+2)+c e^y$
Given, $(x+y+1) \frac{d y}{d x}=1$
$\Rightarrow \quad \frac{d x}{d y}=x+y+1$
$\Rightarrow \quad \frac{d x}{d y}-x=y+1$, which is linear.
$\begin{aligned} & \therefore \quad \text { IF }=e^{\int-1 d y}=e^{-y} \\ & \therefore \text { Solution is } x \cdot e^{-y}=\int(y+1) e^{-y} d y \\ & \Rightarrow x e^{-y}=\int\left(y e^{-y}+e^{-y}\right) d y+c \\ & \Rightarrow x e^{-y}=-y e^{-y}+\int 1 \cdot e^{-y} d y+e^{-y} \cdot(-1)+c \\ & \Rightarrow x e^{-y}=-y e^{-y}-e^{-y}-e^{-y}+c \\ & \Rightarrow x e^{-y}=-(y+2) e^{-y}+c \\ & \Rightarrow \quad x=-(y+2)+c e^y\end{aligned}$
$\Rightarrow \quad \frac{d x}{d y}=x+y+1$
$\Rightarrow \quad \frac{d x}{d y}-x=y+1$, which is linear.
$\begin{aligned} & \therefore \quad \text { IF }=e^{\int-1 d y}=e^{-y} \\ & \therefore \text { Solution is } x \cdot e^{-y}=\int(y+1) e^{-y} d y \\ & \Rightarrow x e^{-y}=\int\left(y e^{-y}+e^{-y}\right) d y+c \\ & \Rightarrow x e^{-y}=-y e^{-y}+\int 1 \cdot e^{-y} d y+e^{-y} \cdot(-1)+c \\ & \Rightarrow x e^{-y}=-y e^{-y}-e^{-y}-e^{-y}+c \\ & \Rightarrow x e^{-y}=-(y+2) e^{-y}+c \\ & \Rightarrow \quad x=-(y+2)+c e^y\end{aligned}$
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