Problem 32 - Calculus for Biology and Medicine

Question

Solve $$ \frac{d^{2} x}{d t^{2}}=-4 x $$ with \(x(0)=0\) and \(x^{\prime}(0)=6\).

Step-by-Step Solution

Answer

The solution is \( x(t) = 3\sin(2t) \).

Step 1: Identify the Type of Equation

This is a second-order linear homogeneous differential equation with constant coefficients: \( \frac{d^{2} x}{dt^{2}} = -4x \). The general approach to solve such equations involves characteristic equations.

Step 2: Form the Characteristic Equation

Assume a solution of the form \( x(t) = e^{rt} \). Plugging this into the differential equation gives the characteristic equation: \( r^2 = -4 \).

Step 3: Solve the Characteristic Equation

The characteristic equation \( r^2 = -4 \) has solutions \( r = \pm 2i \), indicating complex conjugate roots.

Step 4: Write the General Solution

For complex roots \( \pm bi \), the solution takes the form \( x(t) = C_1 \cos(bt) + C_2 \sin(bt) \). Here, \( b = 2 \), so \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) \).

Step 5: Apply Initial Conditions to Find Constants

First, use the condition \( x(0) = 0 \). Substituting gives \( C_1 = 0 \). Next, apply \( x'(0) = 6 \). Calculating the derivative: \( x'(t) = 0 \cdot \cos(2t) + 2C_2 \cdot \cos(2t) = 2C_2 \cdot \cos(2t) \). Substituting \( x'(0) = 6 \) gives \( 2C_2 = 6 \), hence \( C_2 = 3 \).

Step 6: Write the Specific Solution

Now that we have \( C_1 = 0 \) and \( C_2 = 3 \), the specific solution is \( x(t) = 3\sin(2t) \).

Key Concepts

Characteristic Equation

In the realm of solving second-order linear homogeneous differential equations, the characteristic equation plays a pivotal role. These equations typically appear in the form:

• \( \frac{d^2 x}{dt^2} + a \frac{dx}{dt} + bx = 0 \)

To solve these equations, we assume a solution of the form \( x(t) = e^{rt} \), where \( r \) is a constant. This assumption helps transform the differential equation into an algebraic equation known as the characteristic equation. This equation is vital because it allows us to find the correct form of the solution based on the roots of the equation.

• For our problem, the characteristic equation resulting from \( \frac{d^{2} x}{d t^{2}}=-4 x \) is \( r^2 + 4 = 0 \).
• Solving this gives us the possible values of \( r \), which inform the general solution form.

Initial Conditions

Initial conditions are essential as they transform our general solution into a specific one that satisfies the given conditions of a problem. These conditions typically specify the function and its derivatives at a particular point. They are usually given as:

• \( x(0) = x_0 \)
• \( x'(0) = v_0 \)

For our exercise:

• The initial conditions are \( x(0)=0 \) and \( x'(0)=6 \).
• By applying \( x(0)=0 \), we find \( C_1 = 0 \) in the general solution \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) \).
• Using \( x'(0)=6 \) helps us identify \( C_2 = 3 \). With these constants, we narrow down to a specific solution.

By applying these initial conditions correctly, we arrive at \( x(t) = 3\sin(2t) \), which accurately fulfills both conditions laid out in the problem.

Complex Roots

When solving a characteristic equation, the nature of the roots has a direct impact on the solution form:

• Real and distinct roots yield solutions involving exponential functions.
• Real repeated roots result in solutions with polynomials and exponentials.
• Complex conjugate roots introduce trigonometric functions into the solution.

In our scenario, solving \( r^2 = -4 \) yields complex roots: \( r = \pm 2i \). Complex roots always come in conjugate pairs \( a \pm bi \), leading the general solution to take the form:

• \( x(t) = C_1 \cos(bt) + C_2 \sin(bt) \)

This specific structure reflects the periodic nature of solutions when complex roots are involved, corresponding to the sinusoidal functions that emerge from the Euler's formula. Thus, when tackling complex roots, expect to see these trigonometric elements, which offer both a solution method and insights into the system's behavior.