Question

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

• A. \(x+y+1 = c\), where \(c\) is a constant of integration
• B. \(x+y+1 = ce^y\), where \(c\) is a constant of integration
• C. \(x+y+1 = ce^{-y}\), where \(c\) is a constant of integration
• D. \(x+y-1 = ce^{-y}\), where \(c\) is a constant of integration

Hint:

Treat \(x\) as a function of \(y\); the equation becomes linear in \(x\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The equation \((x+y)\frac{dy}{dx} = 1\) is not linear in \(y\), but if we flip it, it becomes linear in \(x\).

Step 2: Rewrite:
\(\frac{dx}{dy} = x + y\), so \(\frac{dx}{dy} - x = y\). This is linear with \(P = -1\) and \(Q = y\).

Step 3: Integrating factor:
\(\text{I.F.} = e^{\int -1\,dy} = e^{-y}\).
\[ x e^{-y} = \int y e^{-y}dy = -ye^{-y} - e^{-y} + c \]

Step 4: Simplify:
Multiply by \(e^{y}\): \(x = -y - 1 + ce^{y}\), so \(x + y + 1 = ce^{y}\).
Option C and D have \(e^{-y}\), which does not come from this integrating factor. Option A is not a family that satisfies the equation.

Final Answer:
The general solution is \(x+y+1 = ce^{y}\), option (B). \[ \boxed{x+y+1 = ce^{y}} \]