Solve the following differential equation e^ dy dx =x+1;y(0)=3 - Vidyadip

Question

Solve the following differential equation
$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1;\text{y}(0)=3$

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Answer

We have
$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$
Taking log on both sides, we get
$\frac{\text{dy}}{\text{dx}}\log\text{e}=\log(\text{x}+1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1) $
$\Rightarrow\text{dy}=\{\log(\text{x}+1)\}\text{dx}$
intregrating both sides ,we get
$\int\text{dy}=\int\{\log(\text{x}+1)\}\text{dx}$
$\Rightarrow\text{y}=\int1\times\log(\text{x}+1)\text{dx}$
$\Rightarrow\text{y}=\log(\text{x}+1)\int1\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x}+1)\int1\text{dx}\Big]\text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\int\frac{\text{x}}{\text{x}+1}\ \text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{y}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}\ ...(1)$
It is given that y(0) = 3
$\therefore3=0\times\log(0+1)-0+\log(0+1)+\text{C}$
$\Rightarrow\text{C}=3$
Substituting the value of C in (1), we get
$\text{y}=\text{x}\log(\text{x}+1)+\log(\text{x}+1)-\text{x}+3$
$\Rightarrow\text{y}=(\text{x}+1)\log(\text{x}+1)-\text{x}+3$
Hence, $\text{y}=(\text{x}+1)\log(\text{x}+1)-\text{x}+3$is the solution to the given differential equation.

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