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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$\big[\text{x}\sqrt{\text{x}^2+\text{y}^2}-\text{y}^2\big]\text{dx}+\text{xy dy}=0$
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Answer

We have,
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}\sqrt{\text{x}^2+\text{y}^2}}{\text{xy}}$
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}\sqrt{\text{x}^2+\text{v}^2\text{x}^2}}{\text{vx}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-\sqrt{1+\text{v}^2}}{\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\frac{\sqrt{1+\text{v}^2}}{\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\sqrt{1+\text{v}^2}}{\text{v}}$
$\Rightarrow\ \frac{\text{v}}{\sqrt{1+\text{v}^2}}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
Putting $1 + v^2 = t$, we get
$\text{v dv}=\frac{\text{dt}}2$
$\therefore\ \frac{1}{2\sqrt{\text{t}}}\text{dt}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{2\sqrt{\text{t}}}\text{dt}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \sqrt{\text{t}}=-\log|\text{x}|+\log\text{C}\ \dots(1)$
Substituting the value of t in (1), we get
$\sqrt{1+\text{v}^2}=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \sqrt{\text{x}^2+\text{y}^2}=\text{x}\log\Big|\frac{\text{C}}{\text{x}}\Big|$
Hence, $\sqrt{\text{x}^2+\text{y}^2}=\text{x}\log\Big|\frac{\text{C}}{\text{x}}\Big|$ is the required solution.

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