Question

The particular solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ when $x = \frac{2}{3}$ and $y = \frac{1}{3}$ is

• A. $2x + 2y - 2 = \log|x+y|$
• B. $y - x + \frac{1}{3} = \log|x+y| $
• C. $x + y - 1 = \log|x+y|$
• D. $4x - 5y - 1 = \log|x+y|$

Hint:

Whenever a differential equation contains a repeating linear group like $(ax+by)$ inside fractions or functions, substituting $v = ax+by$ will immediately decouple the expressions, turning a difficult equation into a straightforward variable-separable integration.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a first-order non-homogeneous differential equation. We need to find its particular solution satisfying the initial boundary conditions $x = \frac{2}{3}$ and $y = \frac{1}{3}$.

Step 2: Key Formula or Approach:
The differential equation contains repeated linear combinations of $(x+y)$ in both the numerator and denominator. We can solve this using the substitution method: $$\text{Put } x + y = v \implies 1 + \frac{dy}{dx} = \frac{dv}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1$$ This substitution transforms the system into a separable differential equation in terms of variables $v$ and $x$.

Step 3: Detailed Explanation:
Substitute these expressions back into the given differential equation: $$\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$$ $$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$$ Separate the variables to prepare for integration: $$\frac{v-1}{2v} \, dv = dx \implies \left(\frac{1}{2} - \frac{1}{2v}\right) dv = dx$$ Integrate both sides of the equation: $$\int \left(\frac{1}{2} - \frac{1}{2v}\right) dv = \int dx$$ $$\frac{v}{2} - \frac{1}{2}\log|v| = x + C$$ Substitute back $v = x + y$: $$\frac{x+y}{2} - \frac{1}{2}\log|x+y| = x + C$$ Multiply the entire equation by 2 to clear out fractions: $$x + y - \log|x+y| = 2x + 2C \implies y - x - 2C = \log|x+y|$$ Let $C' = -2C$, giving our general solution form: $y - x + C' = \log|x+y|$.
Now apply the initial conditions $x = \frac{2}{3}, y = \frac{1}{3}$ to find the particular constant: $$\frac{1}{3} - \frac{2}{3} + C' = \log\left|\frac{2}{3} + \frac{1}{3}\right|$$ $$-\frac{1}{3} + C' = \log(1) = 0 \implies C' = \frac{1}{3}$$ Substitute $C' = \frac{1}{3}$ back into our equation: $$y - x + \frac{1}{3} = \log|x+y|$$

Step 4: Final Answer:
The particular solution is $y - x + \frac{1}{3} = \log|x+y|$, which corresponds to option (B).