Solve the following differential equation: (x+y+1) dy dx = 1 - Vidyadip

Question

Solve the following differential equation:
$(\text{x}+\text{y}+1)\frac{\text{dy}}{\text{dx}} = 1$

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Answer

We have,
$(\text{x}+\text{y}+1)\frac{\text{dy}}{\text{dx}} = 1$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{1}{(\text{x}+\text{y}+1)}$
Let $\text{ x}+\text{y}+1 = \text{v}$
$\therefore1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}-1$
$\therefore\frac{\text{dv}}{\text{dx}}-1=\frac{1}{\text{v}}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{1}{\text{v}}+1$
$\Rightarrow \frac{\text{v}}{\text{v}+1}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}}{\text{v+1}}\text{dv} = \int \text{dx}$
$\Rightarrow \int \frac{\text{v}+1-1}{\text{v}+1}\text{dv}=\int\text{dx}$
$\Rightarrow\int\Big(1-\frac{1}{\text{v}+1}\Big)\text{dv} = \int\text{dx}$
$\Rightarrow\text{v}-\log|\text{v}+1|=\text{x + K}$
$\Rightarrow\text{x + y}+1-\log|\text{x + y}+1+1|=\text{x + K}$
$\Rightarrow\text{y}-\log|\text{x + y}+2|=\text{K}-1$
$\Rightarrow\text{y}-\log|\text{x + y}+2|=\text{C}_1$ $(\text{C}_1 = \text{K}-1)$
$\Rightarrow\text{y}-\text{C}_1=\log|\text{x + y}+2|$
$\Rightarrow\text{e}^{\text{y}-\text{C}_1}=\text{x + y}+2$
$\Rightarrow \frac{\text{e}^\text{y}}{\text{e}^{c_1}} = \text{x}+\text{y}+2$
$\Rightarrow \text{e}^{-\text{c}_1}\text{e}^{\text{y}} = \text{x}+\text{y}+2$
$\Rightarrow \text{Ce}^\text{y} = \text{x}+\text{y}+2$ $(\text{C} = \text{e}^{-\text{c}_1})$
$\Rightarrow \text{x} = \text{Ce}^\text{y} - \text{y}-2$

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