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Differential Equations — Maths STD 12 Science — Question

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

We have,
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{\text{x}^2-\text{xy}+\text{y}^2}$
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2\text{x}^2}{\text{x}^2-\text{vx}^2+\text{v}^2\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}-\text{v}^3}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \frac{1-\text{v}+\text{v}^2}{\text{v}+\text{v}^3}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1+\text{v}^2}{\text{v}(1+\text{v}^2)}\text{dv}-\int\frac{1}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{1+\text{v}^2}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\text{v}|-\tan^{-1}|\text{v}|=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\big|\frac{\text{vx}}{\text{C}}\big|=\tan^{-1}\text{v}$
$\Rightarrow\ \big|\frac{\text{vx}}{\text{C}}\big|=\text{e}^{\tan^{-1}\text{v}}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ |\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$
Hence, $|\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$ is the required solution.

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