Question

Find the general solution of the differential equation \((x^2 - y^2)\,dx + 2xy\,dy = 0\).

Hint:

The equation is homogeneous. Put y = vx, separate the variables and integrate to get x^2 + y^2 = cx.

Answer 1

Step 1: Write the equation in the form dy/dx.
Move the first term across and divide by \(2xy\,dx\).
\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \]
The right side has degree 0 in x and y, so the equation is homogeneous.

Step 2: Substitute y = vx.
Then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).
\[ v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v} \]

Step 3: Separate the variables.
Move v to the right side and combine the terms.
\[ x\frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{1 + v^2}{2v} \]
\[ \frac{2v}{1 + v^2}\,dv = -\frac{dx}{x} \]

Step 4: Integrate both sides.
The left side is of the form \(\int \frac{f^{\prime}(v)}{f(v)}\,dv\), so it gives a logarithm.
\[ \log(1 + v^2) = -\log|x| + \log|c| \]
This gives \((1 + v^2)\,|x| = |c|\), that is \(x(1 + v^2) = c\) after absorbing the sign into c.

Step 5: Put v = y/x back.
\[ x\left(1 + \frac{y^2}{x^2}\right) = c \implies \frac{x^2 + y^2}{x} = c \]
So \(x^2 + y^2 = cx\).

Final Answer:
The general solution, with c an arbitrary constant, is below. \[ \boxed{x^2 + y^2 = cx} \]