Find the particular solution of e^ dy dx =x+1, that y=3, when x=0… - Vidyadip

Question

Find the particular solution of $\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1,$ that $\text{y}=3,$ when $\text{x}=0.$

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Answer

$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$
$\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1),\text{y}=3$ at $\text{x}=0$
$\int\text{dy}=\int\log(\text{x}+1)\text{dx}$
$\text{y}=\log|\text{x}+1|\times\int1\times\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx}+\text{C}$
Using integration by parts
$\text{y = x}\log|\text{x}+1|-\int\frac{\text{x}}{\text{x}+1}\text{dx}+\text{C}$
$\text{y = x}\log|\text{x}+1|-\Big(\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}\Big)+\text{C}$
$=\text{x}\log|\text{x}+1|-(\text{x}-\log|\text{x}+1|)+\text{C}$
$\text{y = x}\log|\text{x}+1|-\text{x}+\log|\text{x}+1|+\text{C}$
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x + C}$
Put $\text{y}=3$ and $\text{x}=0$
$3=0-0+\text{C}$
$\text{C}=3$
Put $\text{C}=3$ in equation (1),
$\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x}+3$

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