Problem 13 - Calculus: Early Transcendentals

Question

Solve the initial value problem \(y^{\prime \prime}+4 y^{\prime}+13 y=0, y(0)=1, y^{\prime}(0)=1\).

Step-by-Step Solution

Answer

The solution is \( y(t) = e^{-2t} ( \cos(3t) + \sin(3t) ) \).

Step 1: Identify the Differential Equation

The differential equation given is a second order linear homogeneous differential equation: \( y'' + 4y' + 13y = 0 \).

Step 2: Find the Characteristic Equation

To solve this differential equation, we first write the characteristic equation, which is obtained by replacing \( y, y', \) and \( y'' \) with \( r^2, r, \) and 1. The characteristic equation is \( r^2 + 4r + 13 = 0 \).

Step 3: Solve the Characteristic Equation

Solve the characteristic equation using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = 4, c = 13 \). Therefore, \( r = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i \).

Step 4: Write the General Solution

Since the roots are complex, the general solution to the differential equation is given by \( y(t) = e^{\alpha t} ( C_1 \cos(\beta t) + C_2 \sin(\beta t) ) \) where \( \alpha = -2 \) and \( \beta = 3 \). Thus, \( y(t) = e^{-2t} ( C_1 \cos(3t) + C_2 \sin(3t) ) \).

Step 5: Apply Initial Conditions

We apply the initial condition \( y(0) = 1 \). Plugging \( t = 0 \) into the general solution: \( y(0) = e^0 ( C_1 \cos(0) + C_2 \sin(0) ) = C_1 = 1 \). Thus, \( C_1 = 1 \).

Step 6: Solve for the Second Constant

Differentiate the general solution to find \( y'(t) \). Using the product and chain rules, we have: \( y'(t) = (e^{-2t})' ( C_1 \cos(3t) + C_2 \sin(3t) ) + e^{-2t} (-(C_1 \sin(3t) + 3C_2 \cos(3t))).\) Evaluating at \( t = 0 \), \( y'(0) = -2(C_1 \cdot 1) + 3C_2 \cdot 1 = -2 + 3C_2 \) Setting \( y'(0) = 1 \), \( -2 + 3C_2 = 1 \) gives \( C_2 = 1 \).

Step 7: Write the Specific Solution

The specific solution to the differential equation is \( y(t) = e^{-2t} ( \cos(3t) + \sin(3t) ) \), using \( C_1 = 1 \) and \( C_2 = 1 \).

Key Concepts

Understanding Differential Equations

A differential equation is an equation that relates a function to its derivatives. In this context, it expresses how a function changes over time or another variable. They're used in various fields like physics, engineering, and economics to model dynamic systems. For instance, the differential equation in our problem is a second-order linear homogeneous equation:
\[ y'' + 4y' + 13y = 0 \]This means both the function \( y(t) \) and its derivatives \( y'(t) \) and \( y''(t) \) are involved, with constant coefficients. The term "homogeneous" means there are no external forces or sources affecting the system. Solving these equations helps us find the function, \( y(t) \), describing the system's behavior over time.

Decoding the Characteristic Equation

To solve a linear differential equation, we often use a related algebraic expression called the characteristic equation. This equation is derived by assuming solutions of the form \( y = e^{rt} \), where \( r \) is a constant. Substituting into the differential equation gives a polynomial equation in terms of \( r \). In our problem:
\[ r^2 + 4r + 13 = 0 \]This characteristic equation helps identify the nature of the roots, which tells us the form of solutions we can expect. Solving it gives us information about whether the system behaves exponentially, sinusoidally, or both.

Exploring Complex Roots

When solving the characteristic equation, you may encounter complex roots, as they are common in oscillatory systems. Using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]gives complex roots if the discriminant \( b^2 - 4ac \) is negative. In our example, the roots are \(-2 \pm 3i\). Complex roots indicate solutions involving exponential decay, oscillation, or a combination:

• Exponential decay: The real part, \(-2\), affects amplitude over time.
• Oscillation: The imaginary part, \(3i\), accounts for the sine and cosine components.

Therefore, the solution has a form \( e^{-2t} (C_1 \cos(3t) + C_2 \sin(3t)) \), capturing both decay and oscillation.

Applying Initial Conditions

Initial conditions are crucial for finding a specific solution to a differential equation. They provide specific values of the function and its derivatives at a particular point in time, such as:

• \( y(0) = 1 \)
• \( y'(0) = 1 \)

These conditions help determine the constants \( C_1 \) and \( C_2 \) in the solution by substituting them into the equation. For instance, using \( y(0) = 1 \) allows us to find \( C_1 = 1 \). Similarly, differentiating the solution and applying \( y'(0) = 1 \) helps find \( C_2 = 1 \). Applying these gives us the tailored specific solution, \( y(t) = e^{-2t} ( \cos(3t) + \sin(3t) ) \), tailored to fit the initial conditions exactly.