Question

The particular solution of differential equation $(x + y) dy + (x - y) dx = 0$ at $x = y = 1$ is

• A. $\log \left| \frac{x^2 + y^2}{2} \right| = \frac{\pi}{2} - 2 \tan^{-1}\left(\frac{y}{x}\right)$
• B. $\log |x^2 + y^2| = \frac{\pi}{2} - 2 \tan^{-1}\left(\frac{y}{x}\right)$
• C. $\log |x^2 + y^2| = \frac{\pi}{2} - 2 \tan^{-1}\left(\frac{y}{x}\right)$
• D. $\log \left| \frac{x^2 + y^2}{2} \right| = \frac{\pi}{4} - 2 \tan^{-1}\left(\frac{y}{x}\right)$

Hint:

To test options quickly in a particular solution question, directly substitute the initial coordinates $x=1, y=1$ into the options! For option (A), the left side becomes $\log(2/2) = \log 1 = 0$, and the right side becomes $\frac{\pi}{2} - 2\tan^{-1}(1) = \frac{\pi}{2} - 2(\frac{\pi}{4}) = 0$. Since $0=0$, option (A) is instantly verified without doing any integration.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a first-order homogeneous differential equation. We need to evaluate its general solution and then substitute the initial conditions $x = 1, y = 1$ to isolate the particular solution.

Step 2: Key Formula or Approach:
1. Express the equation in the standard form: $\frac{dy}{dx} = \frac{y - x}{x + y}$.
2. Since the degrees of all terms are equal, substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$ to separate the variables.
3. Integrate both sides using standard algebraic integration rules.

Step 3: Detailed Explanation:
4. Substitute $y = vx$ into the differential equation: $$v + x\frac{dv}{dx} = \frac{vx - x}{x + vx} = \frac{x(v - 1)}{x(1 + v)} = \frac{v - 1}{1 + v}$$ $$x\frac{dv}{dx} = \frac{v - 1}{1 + v} - v = \frac{v - 1 - v - v^2}{1 + v} = \frac{-(1 + v^2)}{1 + v}$$ 5. Separate the variables $v$ and $x$: $$\frac{1 + v}{1 + v^2} dv = -\frac{1}{x} dx$$ $$\left( \frac{1}{1 + v^2} + \frac{v}{1 + v^2} \right) dv = -\frac{1}{x} dx$$ 6. Integrate both sides: $$\tan^{-1}(v) + \frac{1}{2} \log|1 + v^2| = -\log x + \log c_1$$ Multiply the entire equation by 2 to clear fractions: $$2 \tan^{-1}(v) + \log|1 + v^2| = -2 \log x + \log c$$ Substitute $v = \frac{y}{x}$ back into the equation: $$2 \tan^{-1}\left(\frac{y}{x}\right) + \log\left|1 + \frac{y^2}{x^2}\right| + \log x^2 = \log c$$ $$2 \tan^{-1}\left(\frac{y}{x}\right) + \log\left|\frac{x^2 + y^2}{x^2} \cdot x^2\right| = \log c$$ $$2 \tan^{-1}\left(\frac{y}{x}\right) + \log|x^2 + y^2| = \log c$$ 7. Apply initial conditions $x = 1, y = 1$ to solve for the constant $\log c$: $$2 \tan^{-1}(1) + \log|1^2 + 1^2| = \log c \implies 2\left(\frac{\pi}{4}\right) + \log 2 = \log c \implies \log c = \frac{\pi}{2} + \log 2$$ 8. Substitute $\log c$ back to isolate the particular solution: $$2 \tan^{-1}\left(\frac{y}{x}\right) + \log|x^2 + y^2| = \frac{\pi}{2} + \log 2$$ $$\log|x^2 + y^2| - \log 2 = \frac{\pi}{2} - 2 \tan^{-1}\left(\frac{y}{x}\right)$$ $$\log \left| \frac{x^2 + y^2}{2} \right| = \frac{\pi}{2} - 2 \tan^{-1}\left(\frac{y}{x}\right)$$

Step 4: Final Answer:
The particular solution is given by option (A).