Question

Find a particular solution of the following differential equation:- x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Answer 1

We have,

\[ x^2 dy + \left( xy + y^2 \right)dx = 0\]

\[\frac{dy}{dx} = - \left( \frac{xy + y^2}{x^2} \right)\]

Let y = vx

\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]

\[ \therefore v + x\frac{dv}{dx} = - \left( \frac{v x^2 + v^2 x^2}{x^2} \right)\]

\[ \Rightarrow x\frac{dv}{dx} = - \left( v + v^2 \right) - v\]

\[ \Rightarrow x\frac{dv}{dx} = - 2v - v^2 \]

\[ \Rightarrow \frac{1}{v^2 + 2v}dv = - \frac{1}{x}dx\]

Integrating both sides, we get

\[\int\frac{1}{v^2 + 2v}dy = - \int\frac{1}{x}dx\]

\[\int\frac{1}{v^2 + 2v + 1 - 1}dy = - \int\frac{1}{x}dx\]

\[\int\frac{1}{\left( v + 1 \right)^2 - \left( 1 \right)^2}dy = - \int\frac{1}{x}dx\]

\[ \Rightarrow \frac{1}{2 \times 1}\log\left| \frac{v + 1 - 1}{v + 1 + 1} \right| = - \log \left| x \right| + \log C\]

\[ \Rightarrow \frac{1}{2}\log\left| \frac{v}{v + 2} \right| = - \log \left| x \right| + \log C\]

\[ \Rightarrow \log\left| \frac{v}{v + 2} \right| = - 2\log \left| x \right| + 2\log C\]

\[ \Rightarrow \log\left| \frac{v}{v + 2} \right| + \log \left| x^2 \right| = \log C^2 \]

\[ \Rightarrow \log\left| \frac{v x^2}{v + 2} \right| = \log C^2 \]

\[ \Rightarrow \left| \frac{v x^2}{v + 2} \right| = C^2 \]

\[ \Rightarrow \left| \frac{v x^2}{v + 2} \right| = k,\text{ where }k = 2C\]

\[ \Rightarrow \left| \frac{\frac{y}{x} x^2}{\frac{y}{x} + 2} \right| = k \]

\[ \Rightarrow \left| \frac{x^2 y}{y + 2x} \right| = k . . . . . \left( 1 \right)\]

Now,

When x = 1, y = 1

\[ \therefore \left| \frac{1}{1 + 2} \right| = k \]

\[ \Rightarrow k = \frac{1}{3}\]

Putting the value of k in (1), we get

\[\left| \frac{x^2 y}{y + 2x} \right| = \frac{1}{3}\]

\[ \Rightarrow 3 x^2 y = \pm \left( y + 2x \right)\]

\[\text{But }y\left( 1 \right) = 1\text{ does not satisfy the equation }3 x^2 y = - \left( y + 2x \right)\]

\[ \therefore 3 x^2 y = y + 2x\]

\[ \Rightarrow y = \frac{2x}{3 x^2 - 1}\]