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PDE by separation of variables

Short answer

Assume u = X(x)Y(y); separation gives X'/(x^2 X) = Y'/(y^2 Y) = k. Solving yields X = C1 e^(k x^3/3), Y = C2 e^(k y^3/3). General solution: u(x,y) = C exp(k(x^3 + y^3)/3) with arbitrary constants C and k.

Solve the PDE y²uₓ − x²u_y = 0 by the method of separation of variables.

Assume u(x, y) = X(x)Y(y), where X depends on x only and Y on y only. Then uₓ = X′Y and u_y = XY′.

y² X′ Y − x² X Y′ = 0
Divide by x² y² X Y:     X′/(x² X) = Y′/(y² Y) = k      (constant)

Each side depends on a different variable, so both equal a constant k.

  • X′ = k x² X gives X = C₁ e^(kx³/3)
  • Y′ = k y² Y gives Y = C₂ e^(ky³/3)

Answer: u(x, y) = C exp[ k(x³ + y³)/3 ], where C = C₁C₂ and k are arbitrary constants.

Check: uₓ = kx²u and u_y = ky²u, so y²uₓ − x²u_y = kx²y²u − kx²y²u = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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