Solve the following differential equations: dy dx =1+x+y^2+xy^2 w… - Vidyadip

Question

Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{xy}^2$ when $\text{y}=0,\text{x}=0$

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Answer

$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{xy}^2$
$\frac{\text{dy}}{\text{dx}}=(1+\text{x})(1+\text{y}^2)$
$\frac{1}{(1+\text{y}^2)}\text{dy}=(1+\text{x})\text{dx}$
Integrating on both the sides we get
$\int\frac{1}{(1+\text{y}^2)}\text{dy}=\int(1+\text{x})\text{dx}$
$\tan^{-1}\text{y = x}+\frac{\text{x}^2}{2}+\text{C}...(1)$
Put $\text{y}=0,\text{x}=0$ then
$\tan^{-1}0=0+0+\text{C}$
$\text{C}=0$
From (1) we have
$\tan^{-1}\text{y = x}+\frac{\text{x}^2}{2}$
$\text{y}=\tan\Big(\text{x}+\frac{\text{x}^2}{2}\Big)$

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