Question

Find the general solution of: $$ (2x - y)^2 dy - 2(2x - y)^2 dx - 2 dx = 0 $$

• A. \(\log (2x - y) = 2x + c\)
• B. \((2x - y)^3 + 4 y = c\)
• C. \((2x - y)^3 + 6 x = c\)
• D. \(\log (2x - y) = 2 y + c\)

Hint:

Use substitution and integrate carefully by separating variables.

Answer 1

The Correct Option is C

Rewrite the differential equation: \[ (2x - y)^2 dy - 2(2x - y)^2 dx - 2 dx = 0 \] Divide through by \((2x - y)^2\): \[ dy - 2 dx - \frac{2}{(2x - y)^2} dx = 0 \] Or write as: \[ dy = 2 dx + \frac{2}{(2x - y)^2} dx \] Group terms and integrate both sides or use substitution \( u = 2x - y \). The solution results in: \[ (2x - y)^3 + 6x = c \]