Findx(t),given¨ x=−4xwithinitialconditionsx(t=0)=0, ˙ x(t=0)=1.

The given differential equation is:

\[
\ddot{x} = -4x
\]

where \( \ddot{x} \) represents the second derivative of \( x \) with respect to \( t \), and the initial conditions are \( x(0) = 0 \) and \( \dot{x}(0) = 1 \).

### Step 1: Solve the Homogeneous Equation

The differential equation is a second-order linear homogeneous equation with constant coefficients. The general form is:

\[
\ddot{x} + 4x = 0
\]

This can be rewritten as:

\[
\ddot{x} + 4x = 0
\]

To solve this, assume a solution of the form \( x(t) = e^{rt} \), where \( r \) is a constant. Substituting into the equation:

\[
r^2 e^{rt} + 4e^{rt} = 0
\]

This simplifies to:

\[
e^{rt}(r^2 + 4) = 0
\]

Since \( e^{rt} \) is never zero, we have:

\[
r^2 + 4 = 0
\]

\[
r^2 = -4
\]

\[
r = \pm 2i
\]

### Step 2: Write the General Solution

The general solution to the differential equation is given by:

\[
x(t) = C_1 \cos(2t) + C_2 \sin(2t)
\]

where \( C_1 \) and \( C_2 \) are constants determined by the initial conditions.

### Step 3: Apply Initial Conditions

Given \( x(0) = 0 \) and \( \dot{x}(0) = 1 \), we can find the constants \( C_1 \) and \( C_2 \).

1. **Applying \( x(0) = 0 \):**

\[
x(0) = C_1 \cos(0) + C_2 \sin(0) = C_1(1) + C_2(0) = C_1
\]

So, \( C_1 = 0 \).

2. **Applying \( \dot{x}(0) = 1 \):**

Differentiate \( x(t) \) with respect to \( t \):

\[
\dot{x}(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t)
\]

Substituting \( t = 0 \) and \( \dot{x}(0) = 1 \):

\[
\dot{x}(0) = -2C_1 \sin(0) + 2C_2 \cos(0) = 2C_2
\]

So, \( 2C_2 = 1 \), which gives \( C_2 = \frac{1}{2} \).

### Step 4: Write the Particular Solution

Substituting the values of \( C_1 \) and \( C_2 \) into the general solution:

\[
x(t) = 0 \cdot \cos(2t) + \frac{1}{2} \sin(2t) = \frac{1}{2} \sin(2t)
\]

### Final Answer

The solution to the differential equation with the given initial conditions is:

\[
x(t) = \frac{1}{2} \sin(2t)
\]

Would you like more details or have any questions? Here are some related questions:

1. How can the method of undetermined coefficients be applied to non-homogeneous differential equations?
2. What is the significance of the characteristic equation in solving second-order differential equations?
3. How would the solution change if the initial conditions were different?
4. What is the physical interpretation of this type of differential equation in harmonic motion?
5. Can you explain how to solve second-order differential equations using Laplace transforms?

**Tip:** In problems involving oscillatory motion, the sine and cosine functions often appear as solutions, corresponding to different phases of the motion.

Explore the solution process of a second-order linear homogeneous differential equation with specific initial conditions. Learn step-by-step how to solve differential equations involving harmonic motion.

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