Question

Let \( x + e^x \) and \( 1 + x + e^x \) be solutions of a linear second-order ordinary differential equation with constant coefficients. If \( y(x) \) is the solution of the same equation satisfying \( y(0) = 3 \) and \( y'(0) = 4 \), then \( y(1) \) is equal to

• A. 1
• B. 2e + 3
• C. 3e + 2
• D. 3e + 1

Hint:

When solving a linear differential equation with constant coefficients, express the solution in terms of exponentials, and use the given initial conditions to find the constants.

Answer 1

The Correct Option is D

Step 1: Analyze the given functions.
We are given two solutions of the linear differential equation: \( x + e^x \) and \( 1 + x + e^x \). These solutions suggest that the characteristic equation of the differential equation has roots related to the exponential function, which is a standard feature of constant-coefficient linear differential equations.
Step 2: General solution.
The general solution to the differential equation will be of the form \( y(x) = A e^x + B x e^x \). Use the initial conditions \( y(0) = 3 \) and \( y'(0) = 4 \) to solve for the constants \( A \) and \( B \). \[ y(0) = A + B = 3, \] \[ y'(0) = A + B = 4. \] This system gives \( A = 2 \) and \( B = 1 \). So, the solution is \( y(x) = 2e^x + x e^x \).
Step 3: Find \( y(1) \).
Now substitute \( x = 1 \) into \( y(x) \): \[ y(1) = 2e + e = 3e. \] Thus, the correct answer is \( \boxed{(B)} 2e + 3 \).