Solve the differential equation dy dx =1+x+y^2+xy^2, when y = 0, … - Vidyadip

Question

Solve the differential equation $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{x}\text{y}^2,$ when y = 0, x = 0.

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Answer

Given that, $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^2+\text{x}\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{x})+\text{y}^2(1+\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{y}^2)(1+\text{x})$
$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}=(1+\text{x})\text{dx}$
On integrating both sides, we get
$\tan^{-1}\text{y}=\text{x}+\frac{\text{x}^2}{2}+\text{K}\ .....(\text{i})$
When y = 0 and x = 0. then substituting these values in Eq. (i), we get,
$\tan^{-1}(0)=0+0+\text{K}$
$\Rightarrow\text{K}=0$
$\Rightarrow\tan^{-1}\text{y}=\text{x}+\frac{\text{x}^2}{2}$
$\Rightarrow\text{y}=\tan\Big(\text{x}+\frac{\text{x}^2}{2}\Big)$

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