Solve the differential equation: (x+1) dy dx =2e^ -y -1; y(0)=0. - Vidyadip

Question

Solve the differential equation: $(\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1;\ \text{y(0)}=0.$

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Answer

$(\text{x}+1)\frac{\text{dy}}{\text{dx}}=2\text{e}^{-\text{y}}-1$
$\Rightarrow\frac{\text{dy}}{2\text{e}^{-\text{y}}-1}=\frac{\text{dx}}{\text{x}+1}$
$\Rightarrow\frac{\text{e}^\text{y}\text{dy}}{2-\text{e}^\text{y}}=\frac{\text{dx}}{\text{x}+1}$
Integrating both sides, we get:
$\int\frac{\text{e}^\text{y}\text{dy}}{2-\text{e}^\text{y}}=\log|\text{x}+1|+\log\text{C}\ \dots(1)$
Let $2-\text{e}^\text{y}=\text{t}.$
$\therefore\frac{\text{d}}{\text{dy}}(2-\text{e}^{\text{y}})=\frac{\text{dt}}{\text{dy}}$
$\Rightarrow-\text{e}^\text{y}=\frac{\text{dt}}{\text{dy}}$
$\Rightarrow\text{e}^\text{y}\text{dy}=-\text{dt}$
Substituting this value in equation (1), we get
$\int\frac{-\text{dt}}{\text{t}}=\log|\text{x}+1|+\log\text{C}$
$\Rightarrow-\log|\text{t}|=\log|\text{C}(\text{x}+1)|$
$\Rightarrow-\log|2-\text{e}^\text{y}|=\log|\text{C}(\text{x}+1)|$
$\Rightarrow\frac{1}{2-\text{e}^\text{y}}=\text{C}(\text{x}+1)$
$\Rightarrow2-\text{e}^\text{y}=\frac{1}{\text{C}(\text{x}+1)}\ \dots(2)$
Now, at x = 0 and y = 0, equation (2) becames:
$\Rightarrow2-1=\frac{1}{\text{C}}$
$\Rightarrow\text{c}=1$
Substituting C = 1 in equation (2) we get:
$2-\text{e}^\text{y}=\frac{1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=2-\frac{1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=\frac{2\text{x}+2-1}{\text{x}+1}$
$\Rightarrow\text{e}^\text{y}=\frac{2\text{x}+1}{\text{x}+1}$
$\Rightarrow\text{y}=\log\Big|\frac{2\text{x}+1}{\text{x}+1}\Big|,(\text{x}\neq-1)$
This is the required particular solution of the given differential equaion.

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