Question

Let \( \varphi : (-1, \infty) \to (0, \infty) \) be the solution of the differential equation \[ \frac{dy}{dx} = 2 y e^x = 2 e^x \sqrt{y}, \] satisfying \( \varphi(0) = 1 \). Then, which of the following is/are TRUE?

• A. \( \varphi \) is an unbounded function.
• B. \( \lim_{x \to \ln 2} \varphi(x) = (2e - 1)^2 \).
• C. \( \lim_{x \to \ln 2} \varphi(x) = \sqrt{2e - 1} \).
• D. \( \varphi \) is a strictly increasing function on the interval \( (0, \infty) \).

Hint:

For differential equations involving square roots of \( y \), separating variables and integrating both sides is a common technique to find the solution.

Answer 1

The Correct Option is D

Step 1: Solve the differential equation.
The differential equation is given as: \[ \frac{dy}{dx} = 2 e^x \sqrt{y}. \] Separate the variables: \[ \frac{1}{\sqrt{y}} \, dy = 2 e^x \, dx. \] Integrate both sides: \[ \int \frac{1}{\sqrt{y}} \, dy = \int 2 e^x \, dx. \] This gives: \[ 2 \sqrt{y} = 2 e^x + C. \] Now, solve for \( y \): \[ \sqrt{y} = e^x + \frac{C}{2}, \] \[ y = \left(e^x + \frac{C}{2}\right)^2. \] Step 2: Apply initial condition.
We are given that \( \varphi(0) = 1 \), so: \[ 1 = \left(e^0 + \frac{C}{2}\right)^2. \] This simplifies to: \[ 1 = \left(1 + \frac{C}{2}\right)^2. \] Taking the square root of both sides: \[ 1 + \frac{C}{2} = 1 \quad \Rightarrow \quad \frac{C}{2} = 0 \quad \Rightarrow \quad C = 0. \] Thus, the solution is: \[ \varphi(x) = e^{2x}. \] Step 3: Analyze the function.
- \( \varphi(x) = e^{2x} \) is an exponential function. - It is strictly increasing for \( x>0 \). - As \( x \to \infty \), \( \varphi(x) \to \infty \), so it is unbounded. - Thus, \( \varphi(x) \) is strictly increasing on \( (0, \infty) \). Final Answer: \[ \boxed{\varphi \text{ is a strictly increasing function on the interval } (0, \infty).} \]