Solve the following differential equations: y(1+e^ x )dy=(y+1)e^ … - Vidyadip

Question

Solve the following differential equations:

$\text{y}(1+\text{e}^{\text{x}})\text{dy}=(\text{y}+1)\text{e}^{\text{x}}\text{ dx}$

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Answer

We have,
$\text{y}(1+\text{e}^{\text{x}})\text{dy}=(\text{y}+1)\text{e}^{\text{x}}\text{ dx}$
$\Rightarrow\frac{\text{y}}{\text{y}+1}\text{dy}=\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{y}}{\text{y}+1}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
Substituting $1+\text{e}^{\text{x}}=\text{t},$ we get
$\text{e}^{\text{x}}\text{dx = dt}$
$\therefore\int\frac{\text{y}}{\text{y}+1}\text{dy}=\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\int\frac{\text{y}+1-1}{\text{y}+1}\text{dy}=\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\int\text{dy}-\int\frac{1}{\text{y}+1}\text{dy}=\int\frac{1}{\text{t}}\text{dt}$
$\Rightarrow\text{y}-\log|\text{y}+1|=\log|\text{t}|+\text{C}$
$\Rightarrow\text{y}-\log|\text{y}+1|=\log|1+\text{e}^{\text{x}}|+\text{C}$

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