Question

The solution of the linear differential equation \[ \frac{dy}{dx} + y = e^x, \] when \( y(0) = 0 \), is:

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

Hint:

For first-order linear differential equations, the integrating factor method is an effective technique for finding the solution. Remember to apply initial conditions to find the constant.

Answer 1

The Correct Option is A

This is a first-order linear differential equation. We solve it using the integrating factor method. The equation is: \[ \frac{dy}{dx} + y = e^x. \] The integrating factor is given by: \[ \mu(x) = e^{\int 1 \, dx} = e^x. \] Multiplying both sides of the differential equation by the integrating factor \( e^x \), we get: \[ e^x \frac{dy}{dx} + e^x y = e^{2x}. \] The left-hand side is the derivative of \( y e^x \), so we can rewrite the equation as: \[ \frac{d}{dx}(y e^x) = e^{2x}. \] Now, integrate both sides with respect to \( x \): \[ y e^x = \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C. \] Thus, \[ y = \frac{1}{2} e^x + C e^{-x}. \] Using the initial condition \( y(0) = 0 \): \[ 0 = \frac{1}{2} e^0 + C e^0 \quad \Rightarrow \quad C = -\frac{1}{2}. \] Therefore, the solution is: \[ y = \frac{1}{2} e^x - \frac{1}{2} e^{-x}. \]