Question

Consider the following differential equation \[ (1 + y) \frac{dy}{dx} = y. \] The solution of the equation that satisfies the condition \( y(1) = 1 \) is

• A. \( 2y e^y = e^x + e \)
• B. \( y^2 e^y = e^x \)
• C. \( y e^y = e^x \)
• D. \( (1 + y) e^y = 2 e^x \)

Hint:

For separable differential equations, rearrange terms to separate the variables, integrate both sides, and apply the initial condition to find the constant of integration.

Answer 1

The Correct Option is C

We are given the differential equation: \[ (1 + y) \frac{dy}{dx} = y. \] Step 1: Rearrange the equation Rearrange the equation as: \[ \frac{dy}{dx} = \frac{y}{1 + y}. \] Step 2: Separate variables Separate the variables: \[ \frac{1 + y}{y} \, dy = dx. \] This simplifies to: \[ \left( 1 + \frac{1}{y} \right) dy = dx. \] Step 3: Integrate both sides Now, integrate both sides: \[ \int \left( 1 + \frac{1}{y} \right) dy = \int dx. \] On integrating: \[ y + \ln |y| = x + C, \] where \( C \) is the constant of integration. Step 4: Apply the initial condition We are given the initial condition \( y(1) = 1 \). Substituting this into the equation: \[ 1 + \ln 1 = 1 + C $\Rightarrow$ 1 + 0 = 1 + C $\Rightarrow$ C = 0. \] Thus, the solution is: \[ y + \ln |y| = x. \] Exponentiate both sides: \[ y e^y = e^x. \] Thus, the solution is \( y e^y = e^x \), corresponding to Option (C).
Final Answer: (C) \( y e^y = e^x \)