Problem 10 - Calculus: Early Transcendentals

Question

Solve the initial value problem \(y^{\prime \prime}+6 y^{\prime}+18 y=0, y(0)=0, y^{\prime}(0)=6\).

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Differential Equation

The given problem is a second-order linear homogeneous differential equation with constant coefficients: \(y^{\prime \prime}+6 y^{\prime}+18 y=0\), along with initial conditions \(y(0)=0\) and \(y^{\prime}(0)=6\).

Step 2: Write the Characteristic Equation

To solve the differential equation, start by finding the characteristic equation. For an equation of the form \(a y^{\prime \prime} + b y^{\prime} + c y = 0\), the characteristic equation is \(a r^2 + b r + c = 0\). Substitute here: \(r^2 + 6r + 18 = 0\).

Step 3: Solve the Characteristic Equation

Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve \(r^2 + 6r + 18 = 0\). Compute the discriminant: \(b^2 - 4ac = 6^2 - 4 \times 1 \times 18 = 36 - 72 = -36\). Since the discriminant is negative, the roots are complex. The roots are \(r = \frac{-6 \pm i\sqrt{36}}{2} = -3 \pm 3i\).

Step 4: Write the General Solution

With complex roots \( \alpha \pm \beta i\) where \( \alpha = -3\) and \( \beta = 3\), the general solution of the differential equation is \(y(t) = e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t))\). Thus, the solution becomes \(y(t) = e^{-3t}(C_1\cos(3t) + C_2\sin(3t))\).

Step 5: Apply Initial Conditions

Use the initial conditions to find \(C_1\) and \(C_2\). Start with \(y(0) = 0\): \(e^{0}(C_1\cos(0) + C_2\sin(0)) = 0 \Rightarrow C_1 = 0\). Now calculate \(y^{\prime}(t)\) and use \(y^{\prime}(0) = 6\). Derive \(y^{\prime}(t) = -3e^{-3t}C_2\sin(3t) + e^{-3t}(3C_2\cos(3t))\). Evaluate at \(t=0\): \(-3C_2 + 3C_2 = 6\) which simplifies, solving \(C_2 = 2\).

Step 6: Construct the Solution with Constants

Input the calculated constants \(C_1 = 0\) and \(C_2 = 2\) into the general solution. Thus, \(y(t) = e^{-3t}(0\cdot \cos(3t) + 2\cdot \sin(3t)) = 2e^{-3t}\sin(3t)\).

Key Concepts

Characteristic Equation

A key step in solving second-order linear homogeneous differential equations is constructing the characteristic equation. This equation helps us find the roots—which are essential for determining the solution to the differential equation. In problems like these, the setup is usually in the form

• \(a y^{\prime \prime} + b y^{\prime} + c y = 0\)

To construct the characteristic equation, we substitute the differential equation with a polynomial of the form

• \(a r^2 + b r + c = 0\)

These coefficients \(a\), \(b\), and \(c\) simply correspond to the coefficients of the original differential equation. For our exercise, substituting in the numbers gives the characteristic equation \(r^2 + 6r + 18 = 0\).

Solving this characteristic equation is crucial because it provides us the roots, \(r\), needed to form the general solution for the differential equation. Understanding and setting up the characteristic equation is fundamental to progression through these exercises.

Complex Roots

When analyzing the characteristic equation, the nature of its roots—whether real or complex—directly influences the form of the solution. In our problem, the characteristic equation \(r^2 + 6r + 18 = 0\) is found to have complex roots. This occurs because the discriminant, given by \(b^2 - 4ac\), is negative. Here,

• \(b^2 - 4ac = 36 - 72 = -36\)

Therefore, the roots come out as \(r = -3 \pm 3i\). Complex roots are typically written in the form \(\alpha \pm \beta i\), where \(\alpha\) is the real part and \(\beta\) is the imaginary part.

The general solution with complex roots integrates the concept of exponential decay with sinusoidal oscillation:

• \(y(t) = e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t))\)

This results in solutions that behave differently over time compared to those with real roots, often showing oscillating behavior which is damped. Complex roots are thus crucial in capturing dynamics of systems described by such differential equations.

Initial Value Problem

An initial value problem not only requires finding the general solution of a differential equation but also necessitates using specific conditions to precisely determine the constants within that solution. In our example, the initial conditions given are \(y(0)=0\) and \(y^{\prime}(0)=6\). These provide the necessary information to solve for the coefficients \(C_1\) and \(C_2\) in the general solution.

Applying \(y(0)=0\), we discover:

• \(e^{0}(C_1\cos(0) + C_2\sin(0)) = 0 \Rightarrow C_1 = 0\)

Additionally, with \(y^{\prime}(0)=6\), we find that the constants meet the relationship \(-3C_2 + 3C_2 = 6\), leading us to \(C_2 = 2\).

Ultimately, substituting \(C_1\) and \(C_2\) back into the general solution, we obtain \(y(t) = 2e^{-3t}\sin(3t)\). Solving initial value problems bridges theoretical general solutions to practical specifics, vital for applications in real-world scenarios.