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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The solution of the differention equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}$ is:
  • A
    $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$
  • ✓
    $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
  • C
    $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
  • D
    $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$
✓

Answer

Correct option: B.
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}\ ...(\text{i})$
This is homogenous differential equation.
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
Now, putting equation (i),
$\text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\frac{\text{x}^{2}+\text{x}^{2}\text{u}+\text{x}^{2}\text{u}^{2}}{\text{x}^{2}}$
$\Rightarrow \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}+\text{u}^{2}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}^{2}$
$\Rightarrow \big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\frac{1}{\text{x}}\text{dx}$
Intergreting both sides, we get
$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$
$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$

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