Question

The general solution of the differential equation $(2y - 1) \, dx - (2x + 3) \, dy = 0$ is

• A. $(2x + 3)^2 = c(2y - 1)$
• B. $\frac{2x + 3}{2y - 1} = c$
• C. $(2x + 3)(2y - 1) = c$
• D. $(2x + 3)(2y - 1)^2 = c$

Hint:

When integrating terms where both variables have identical linear coefficients (like $2x$ and $2y$), their integration factors ($\frac{1}{2}$) cancel each other out completely. You can group them directly inside a single fraction constant: $\frac{\text{linear group}_1}{\text{linear group}_2} = c$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a first-order differential equation $(2y - 1) \, dx - (2x + 3) \, dy = 0$. We need to find its general mathematical solution.

Step 2: Key Formula or Approach:
The equation can be solved by grouping the terms using the method of separation of variables: $$M(x) \, dx + N(y) \, dy = 0 \implies \int M(x) \, dx + \int N(y) \, dy = C$$ Recall the standard logarithmic integration rule: $\int \frac{1}{ax + b} \, dx = \frac{1}{a} \log|ax + b|$.

Step 3: Detailed Explanation:
Rearrange the differential equation to group matching variables on opposite sides: $$(2y - 1) \, dx = (2x + 3) \, dy$$ Divide both sides by $(2y - 1)(2x + 3)$ to isolate the variables: $$\frac{dx}{2x + 3} = \frac{dy}{2y - 1}$$ Integrate both sides of the equation: $$\int \frac{1}{2x + 3} \, dx = \int \frac{1}{2y - 1} \, dy$$ $$\frac{\log|2x + 3|}{2} = \frac{\log|2y - 1|}{2} + c_1$$ Multiply the entire equation by 2 to clear out the denominators: $$\log|2x + 3| = \log|2y - 1| + 2c_1$$ Let $2c_1 = \log c$, where $c$ is a new arbitrary constant: $$\log|2x + 3| - \log|2y - 1| = \log c$$ Apply the logarithmic quotient rule $\log A - \log B = \log\left(\frac{A}{B}\right)$: $$\log\left|\frac{2x + 3}{2y - 1}\right| = \log c$$ Taking the antilogarithm of both sides yields the final structural general form: $$\frac{2x + 3}{2y - 1} = c$$ This matches option (B).

Step 4: Final Answer:
The general solution of the differential equation is $\frac{2x + 3}{2y - 1} = c$, which corresponds to option (B).