Question

Find the particular solution of the differential equation:
\[ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \] given that \( y = 1 \) when \( x = 0 \).

Answer 1

Step 1: Try substitution:
$\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.
Divide numerator and denominator by $y^2$:
$\frac{1}{y} \frac{dy}{dx} = \frac{x}{x^2 + y^2}$.
But easier: Try substitution $y = vx$.

Step 2: Let $y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substitute:
$v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + v^2x^2} = \frac{v}{1 + v^2}$.

Step 3: So,
$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v(1 - (1 + v^2))}{1 + v^2} = \frac{-v^3}{1 + v^2}$.

Step 4: Separate variables:
$\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$.

Step 5: Integrate:
$\int \left( \frac{1}{v^3} + \frac{1}{v} \right) dv = -\ln|x| + C$.
$\int v^{-3} dv + \int v^{-1} dv = -\ln|x| + C$.
$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.

Step 6: Use initial condition: when $x = 0$, $y = 1$.
So, $y = vx \Rightarrow 1 = v \cdot 0 \Rightarrow$ undefined.
Better: separate original equation.

Alternate approach:
Try $y^2 = z \Rightarrow y = \sqrt{z} \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{z}} \cdot \frac{dz}{dx}$.
This is a Bernoulli-type differential equation.
Try to solve directly: Not feasible here due to initial condition — alternate methods are better.