Question

The general solution of the differential equation \[ (y^2 + x + 1) dy = (y + 1) dx \] is: 

• A. \( x + 2 + (y+1) \log(y+1)^2 = y + c \)
• B. \( x + 2 + \log(y+1)^2 = \frac{y}{y+1} + c \)
• C. \( \frac{x}{y+1} = \log(y+1)^2 + y + c \)
• D. \( \frac{x+2}{y+1} \log(y+1)^2 = y + c \)

Hint:

To solve first-order differential equations, try expressing them in a separable form. Then integrate both sides accordingly.

Answer 1

The Correct Option is D

Step 1: Rewrite the given equation
The given differential equation is: \[ (y^2 + x + 1) dy = (y + 1) dx. \] Rearrange it as: \[ \frac{dy}{dx} = \frac{y+1}{y^2 + x + 1}. \] Step 2: Use variable separable method
Rewriting, \[ \frac{y^2 + x + 1}{y+1} dy = dx. \] Separate the variables: \[ \left( \frac{y^2}{y+1} + \frac{x+1}{y+1} \right) dy = dx. \] Integrating both sides: \[ \int \left( y - 1 + \frac{x+2}{y+1} \right) dy = \int dx. \] Step 3: Solve the integral
Solving, \[ \frac{x+2}{y+1} \log(y+1)^2 = y + c. \] Step 4: Final Answer
Thus, the general solution of the given differential equation is: \[ \boxed{\frac{x+2}{y+1} \log(y+1)^2 = y + c}. \]