Question

The solution of the equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{x+y+1}\) is

• A. \(x=\mathrm{log}(x+y+2)+\mathrm{c}\), where c is the constant of integration
• B. \(x=\mathrm{log}(x+y-2)+\mathrm{c}\), where c is the constant of integration
• C. \(y=\mathrm{log}(x+y+2)+c\), where \(c\) is the constant of integration
• D. \(y=\mathrm{log}(x+y-2)+\mathrm{c}\), where c is the constant of integration

Hint:

Invert the equation to get a linear equation for x as a function of y.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Take the reciprocal: \(\dfrac{dx}{dy}=x+y+1\), which is linear in \(x\).

Step 2: Integrating factor
\(\dfrac{dx}{dy}-x=y+1\) has integrating factor \(e^{-y}\).

Step 3: Integrate
\(x e^{-y}=\int(y+1)e^{-y}dy=-(y+2)e^{-y}+C\).

Step 4: Calculation
\(x=-(y+2)+Ce^{y}\), so \(x+y+2=Ce^{y}\). Taking logarithm: \(y=\log(x+y+2)-\log C\), i.e. \(y=\log(x+y+2)+c\).

Final Answer:
The solution is \(y=\log(x+y+2)+c\). \[ \boxed{\text{(C) }y=\log(x+y+2)+c} \]