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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
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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