Question

The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is –

• A. log |2x + y – 1| = C + x + y
• B. log (4x + 2y – 1) = C + 2x + y
• C. log (2x + y + 1) + x + 2y = C
• D. log |2x + y – 1| + x + 2y = C

Answer 1

Answer: (D)

log |2x + y – 1| + x + 2y = C

Answer 2

Put 2x + y = X Ž 2 + $\frac{dy}{dx}$ = $\frac{dX}{dx}$. Therefore, the given equation is reduced to (see theory the case when aB – bA = 0)

$\frac{dX}{dx}$ – 2 = – $\frac{X + 1}{2X - 1}$ Ž $\frac{dX}{dx}$ = $\frac{3(X - 1)}{2X - 1}$

Ž $\frac{2X - 1}{3(X - 1)}$ dX = dx Ž $\frac{1}{3}$ $\left\lbrack 2 + \frac{1}{X - 1} \right\rbrack$ dX = dx

Ž $\frac{1}{3}$ [2X + log |X – 1|] = x + const

Ž 2(2x + y) + log |2x + y – 1| = 3x + const

Ž x + 2y + log |2x + y – 1| = C.