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Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following differential equation:
$\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$
✓

Answer

$\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{x}^2\text{v}+\text{v}^2\text{x}^2}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=1+\text{v}+{\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\big(1+{\text{v}^2}\big)$
$\Rightarrow\ \frac{1}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \tan^{-1}{\text{v}}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{x}|+\text{C}$
Hence, $\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{x}|+\text{C}$ is the required solution.

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