Question

The solution of the equation \((\frac{dy}{dx} = \frac{1}{x+y+1})\) is

• A. \((x = \log(x + y + 2) + c)   \)
• B. \((x = \log(x + y - 2) + c) \)
• C. \((y = \log(x + y + 2) + c) \)
• D. \((y = \log(x + y - 2) + c)\)

Hint:

If $\frac{dy}{dx}$ is hard to solve, try solving for $\frac{dx}{dy}$.

Answer 1

The Correct Option is C

Step 1: Reciprocate the equation 
\((\frac{dx}{dy} = x + y + 1 \implies \frac{dx}{dy} - x = y + 1). \)
Step 2: Linear Differential Equation in (x) 
\((P = -1, Q = y + 1). \)
\((I.F. = e^{\int -1 dy} = e^{-y}). \)
Step 3: Solve 
\((x \cdot e^{-y} = \int (y+1) e^{-y} , dy).  Using parts: (\int (y+1) e^{-y} , dy = -(y+1)e^{-y} - e^{-y} = -e^{-y}(y+2)).  (x \cdot e^{-y} = -e^{-y}(y+2) + C \implies x = -(y+2) + Ce^y).  Rearranging to match options: (x+y+2 = Ce^y \implies y = \ln(x+y+2) + c). \)
Final Answer: (C)