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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
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Answer

We have,
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+3\text{xy}+\text{y}^2}{\text{x}^2}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+3\text{vx}^2+\text{v}^2\text{x}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+3\text{v}+\text{v}^2-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2+2\text{v}$
$\Rightarrow\ \frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{(1+\text{v})^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\frac{1}{(1+\text{v})}=\log|\text{x}|+\text{C}$
$\Rightarrow\ \log|\text{x}|+\frac{1}{(1+\text{v})}=-\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$
where
$C_1 = -C$
Hence, $\log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$ is the required solution.

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