Problem 33 - Calculus for Biology and Medicine

Question

Solve $$ \frac{d^{2} x}{d t^{2}}=-9 x $$ with \(x(0)=0\) and \(x^{\prime}(0)=12\).

Step-by-Step Solution

Answer

The solution is \( x(t) = 4 \sin(3t) \).

Step 1: Identify the Type of Equation

The given equation is a second order linear homogeneous differential equation: \( \frac{d^{2} x}{d t^{2}} = -9x \). Our goal is to find the general solution to this equation.

Step 2: Assume a Solution Form

We assume a solution of the form \( x(t) = e^{r t} \). This is a standard approach for solving linear differential equations with constant coefficients.

Step 3: Substitute and Simplify

Substitute \( x(t) = e^{r t} \) into the equation to find the characteristic equation: \( r^2 e^{r t} = -9 e^{r t} \). Simplifying, we get the characteristic equation: \( r^2 + 9 = 0 \).

Step 4: Solve Characteristic Equation

Solve the characteristic equation \( r^2 + 9 = 0 \). The solutions are \( r = \pm 3i \), giving us complex roots.

Step 5: Write the General Solution

With complex roots \( r = \pm 3i \), the general solution is cosine and sine functions: \( x(t) = C_1 \cos(3t) + C_2 \sin(3t) \).

Step 6: Apply Initial Conditions

First apply \( x(0) = 0 \): \( C_1 \cos(0) + C_2 \sin(0) = 0 \), simplifying to \( C_1 = 0 \). Now apply \( x'(0) = 12 \) knowing \( x'(t) = -3 C_1 \sin(3t) + 3C_2 \cos(3t) \). Solve \( 3C_2 = 12 \) to get \( C_2 = 4 \).

Step 7: Final Solution

Substitute \( C_1 = 0 \) and \( C_2 = 4 \) into the general solution. Thus, \( x(t) = 4 \sin(3t) \) is the specific solution satisfying the initial conditions.

Key Concepts

Linear Homogeneous Differential Equations

A linear homogeneous differential equation is a special type of differential equation that plays a critical role in understanding oscillatory behavior in systems. In general, such equations have the form: \[ a_n \frac{d^n x}{dt^n} + a_{n-1} \frac{d^{n-1} x}{dt^{n-1}} + \ldots + a_1 \frac{dx}{dt} + a_0 x = 0 \]where all terms involve the function or its derivatives, and they all add up to zero.
- **Linear:** It means that the dependent variable and all its derivatives appear to the power of one. There are no products of the variables or their derivatives.- **Homogeneous:** This term indicates that the right-hand side of the equation is zero.These equations are important because they can often be solved using characteristic equations, which are derived from assuming solutions of a particular form. This sets a probe to analyze how small oscillations or damped behaviors resolve themselves in many systems.

Complex Roots

In our second-order differential equation, solving the characteristic equation yielded complex roots: \( r = \pm 3i \). Complex roots arise when the coefficients of the differential equation lead to a negative discriminant in the quadratic solution process. This is indicative of oscillatory solutions without real exponential growth or decay. The complex roots suggest the solution involves harmonic functions like sine and cosine.
- **General Solution Form:** For any second order linear homogeneous differential equation with complex roots \( r = a \pm bi \), the general solution is: \[ x(t) = e^{at} (C_1 \cos(bt) + C_2 \sin(bt)) \] When the real part, \( a \), is zero, the solution simplifies to: \[ x(t) = C_1 \cos(bt) + C_2 \sin(bt) \] In our example, \( a = 0 \) and \( b = 3 \), thus leading to: \[ x(t) = C_1 \cos(3t) + C_2 \sin(3t) \] emphasizing the role of trigonometric functions in capturing the dynamic behaviors.

Initial Conditions

Initial conditions are crucial in determining the specific solution to a differential equation among the infinitely many possible solutions that satisfy the general form. These conditions are given as values at a starting point (often when \( t = 0 \)) and help identify the constants in the general solution:- **Example:** Given that \( x(0) = 0 \) and \( x'(0) = 12 \), these initial conditions allow us to solve for \( C_1 \) and \( C_2 \) in the equation \( x(t) = C_1 \cos(3t) + C_2 \sin(3t) \).Upon applying these:- From \( x(0) = 0 \), we find \( C_1 = 0 \).- From \( x'(0) = 12 \), solving gives \( C_2 = 4 \).Thus, the specific solution that adheres to these initial conditions is:\[ x(t) = 4 \sin(3t) \]This illustrates how initial conditions tailor a broad set of solutions to fit particular beginning states of a physical system.