The solution of the differential equation dy dx =1+x+y^ 2 +xy^ 2 … - Vidyadip

MCQ

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}, \text{y}=(0)$ is :

A
$\text{y}^{2}=\text{exp}\big(\text{x}+\frac{\text{x}^{2}}{2}-1\big)$
B
$\text{y}^{2}=1+\text{C}\ \text{exp}\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$
C
$\text{y}=\tan (\text{C}+\text{x}+\text{x}^{2})$
✓
$\text{y}=\tan\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$

✓

Answer

Correct option: D.

$\text{y}=\tan\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$

We have,
$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)\text{y}^{2}(\text{x}+1)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)(1+\text{y})$
$\Rightarrow \frac{\text{dy}}{(1+\text{y}^{2})}=(\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int \frac{\text{dy}}{(1+\text{y}^{2})}=\int(\text{x}+1)\text{dx}$
$\Rightarrow \tan^{-1}=\frac{\text{x}^{2}}{2}+\text{x}+\text{C}\ ...(\text{i})$
Now, $y(0) = 0$
$\therefore\ \tan^{-1}(0)=\frac{\text{0}}{2}+\text{0}+\text{C}$
$\Rightarrow \text{C}=0$
Putting the value of $C$ in $(i),$
$\Rightarrow \tan^{-1}\text{y}=\frac{\text{x}^{2}}{2}+\text{x}$
$\Rightarrow \text{y}=\tan\big(\frac{\text{x}^{2}}{2}+\text{x}\big)$

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