Question

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$ is

• A. $3(x+y) + 4\log|3x+6y-1| = K$
• B. $3(x-y) + 4\log|3x+6y-1| = K$
• C. $6(-x+y) + 4\log|3x+6y-1| = K$
• D. $6(x+y) + 4\log|3x+6y-1| = K$

Hint:

When $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, substitution of $ax+by$ is always the most effective path to convert the equation into variables separable form!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Solve the differential equation where the numerator and denominator are linear functions of $x$ and $y$.

Step 2: Key Formula or Approach:
Use the substitution $t = x + 2y$. Then $dt = dx + 2dy \implies dy = (dt - dx)/2$.

Step 3: Detailed Explanation:
$\frac{dy}{dx} = \frac{t-1}{t+1} \implies \frac{1}{2}(\frac{dt}{dx} - 1) = \frac{t-1}{t+1} \implies \frac{dt}{dx} = \frac{2t-2}{t+1} + 1 = \frac{3t-1}{t+1}$.
$\int \frac{t+1}{3t-1} dt = \int dx \implies \frac{1}{3} \int \frac{3t+3}{3t-1} dt = x + c \implies \frac{1}{3} \int (1 + \frac{4}{3t-1}) dt = x + c$.
$t/3 + 4/9 \log|3t-1| = x + c \implies t + 4/3 \log|3t-1| = 3x + K$.
$x+2y + 4/3 \log|3(x+2y)-1| = 3x + K \implies 2y - 2x + 4/3 \log|3x+6y-1| = K$.
Multiply by 3: $6(y-x) + 4 \log|3x+6y-1| = K$.

Step 4: Final Answer:
The general solution is $6(-x+y) + 4\log|3x+6y-1| = K$, matching option (C).