Question

Find the general solution of the differential equation \( (x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy \).

Hint:

• Homogeneous equations are identified by having the same total degree for every term.
• Always substitute \( y=vx \) and then look for separation of variables.

Answer 1

Concept:
• Homogeneous differential equation of degree 3.
• Substitution: \( y = vx, \frac{dy}{dx} = v + x \frac{dv}{dx} \).

Step 1: Express the equation in \( \frac{dy}{dx} \) form
\( \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \)

Step 2: Substitute \( y = vx \)
\( v + x \frac{dv}{dx} = \frac{x^3 - 3x(vx)^2}{(vx)^3 - 3x^2(vx)} = \frac{x^3(1 - 3v^2)}{x^3(v^3 - 3v)} = \frac{1 - 3v^2}{v^3 - 3v} \)
\( x \frac{dv}{dx} = \frac{1 - 3v^2}{v^3 - 3v} - v = \frac{1 - 3v^2 - v^4 + 3v^2}{v^3 - 3v} = \frac{1 - v^4}{v^3 - 3v} \)

Step 3: Separate variables and integrate
\( \int \frac{v^3 - 3v}{1 - v^4} dv = \int \frac{1}{x} dx \)
Let \( v^2 = t \implies 2v dv = dt \):
\( \frac{1}{2} \int \frac{t - 3}{1 - t^2} dt = \int \frac{1}{x} dx \)
Using partial fractions on \( \frac{t-3}{(1-t)(1+t)} \):
\( \frac{1}{2} \int \left( \frac{-1}{1-t} - \frac{2}{1+t} \right) dt = \ln|x| + C' \)
\( \frac{1}{2} [ \ln|1-t| - 2 \ln|1+t| ] = \ln|x| + C' \)
\( \ln \left( \frac{\sqrt{1-v^2}}{1+v^2} \right) = \ln|kx| \)

Step 4: Substitute back \( v = y/x \)
Simplifying leads to the relation:
\( x^2 - y^2 = C(x^2 + y^2)^2 \).