Math Problem Statement

Consider a system model given by

fraction numerator d squared x left parenthesis t right parenthesis over denominator d t squared end fraction equals t x left parenthesis 0 right parenthesis equals 1 right enclose fraction numerator d x left parenthesis t right parenthesis over denominator d t end fraction end enclose subscript t equals 0 end subscript equals 2.

What is the forced response?

1 over 6 t cubed plus 2 t plus 1

2 t plus 1

1

1 over 6 t cubed plus 2 t

1 over 6 t cubed

2 t cubed plus 3 t plus 1

Solution

We are given the second-order differential equation:

d2x(t)dt2=t\frac{d^2x(t)}{dt^2} = t

with the initial conditions:

x(0)=1anddx(t)dtt=0=2.x(0) = 1 \quad \text{and} \quad \frac{dx(t)}{dt}\bigg|_{t=0} = 2.

Step 1: Solve the differential equation

To solve this, we need to integrate the equation twice to find x(t)x(t).

First integration

d2x(t)dt2=t\frac{d^2x(t)}{dt^2} = t Integrating both sides with respect to tt: dx(t)dt=tdt=t22+C1,\frac{dx(t)}{dt} = \int t \, dt = \frac{t^2}{2} + C_1, where C1C_1 is a constant of integration.

Second integration

Now, integrate again to find x(t)x(t): x(t)=(t22+C1)dt=t36+C1t+C2,x(t) = \int \left(\frac{t^2}{2} + C_1\right) dt = \frac{t^3}{6} + C_1t + C_2, where C2C_2 is another constant of integration.

Step 2: Apply the initial conditions

Initial condition 1: x(0)=1x(0) = 1

Substitute t=0t = 0 into the expression for x(t)x(t): x(0)=036+C1(0)+C2=C2=1.x(0) = \frac{0^3}{6} + C_1(0) + C_2 = C_2 = 1. Thus, C2=1C_2 = 1.

Initial condition 2: dx(t)dtt=0=2\frac{dx(t)}{dt}\bigg|_{t=0} = 2

We already have dx(t)dt=t22+C1\frac{dx(t)}{dt} = \frac{t^2}{2} + C_1. Substitute t=0t = 0: dx(t)dtt=0=022+C1=C1=2.\frac{dx(t)}{dt}\bigg|_{t=0} = \frac{0^2}{2} + C_1 = C_1 = 2. Thus, C1=2C_1 = 2.

Step 3: Final solution

The solution for x(t)x(t) is: x(t)=t36+2t+1.x(t) = \frac{t^3}{6} + 2t + 1.

This matches the first option:

t36+2t+1.\boxed{\frac{t^3}{6} + 2t + 1}.

Would you like more details or have any questions?

Here are five related questions:

  1. What is the homogeneous solution for this system?
  2. How would the solution change if the initial conditions were different?
  3. What is the physical meaning of a forced response in this context?
  4. Can you solve this problem using Laplace transforms?
  5. How does the second derivative in this system affect the shape of the solution?

Tip: When solving differential equations, always remember to check if it's homogeneous or non-homogeneous, as that determines the form of the solution.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Initial Value Problems

Formulas

d^2x(t)/dt^2 = t
dx(t)/dt = ∫t dt = t^2/2 + C₁
x(t) = ∫(t^2/2 + C₁) dt = t^3/6 + C₁t + C₂

Theorems

Integration of functions
Initial Conditions

Suitable Grade Level

Grades 11-12