Problem 36 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$y^{\prime \prime}-4 y^{\prime}+5 y=0, \quad y(0)=3, \quad y^{\prime}(0)=5$$

Step-by-Step Solution

Answer

The specific solution for the given initial-value problem is \(y(t) = e^{2t}(3 \cos(t) - \sin(t))\).

Step 1: Find the characteristic equation

We will start by finding the characteristic equation of the given differential equation. The characteristic equation has the form \(r^2 - 4r + 5 = 0\).

Step 2: Solve the characteristic equation

To find the roots of the characteristic equation, we can either use the quadratic formula or factoring. In this case, we can't easily factor the equation, so we will use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where a=1, b=-4, and c=5. Plugging in these values, we get: \(r = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2}\) \(r = \frac{4 \pm \sqrt{16 - 20}}{2}\) \(r = \frac{4 \pm \sqrt{-4}}{2}\) \(r = 2 \pm i\) So, the roots of the characteristic equation are complex: \(r_1 = 2 + i\) and \(r_2 = 2 - i\).

Step 3: Write the general solution

Since we have complex roots, the general solution of the differential equation takes the form: \(y(t) = e^{at}(c_1 \cos(bt) + c_2 \sin(bt))\) where \(a\) is the real part of the roots, and \(b\) is the imaginary part of the roots. In our case, \(a = 2\) and \(b = 1\). So, the general solution is: \(y(t) = e^{2t}(c_1 \cos(t) + c_2 \sin(t))\)

Step 4: Apply the initial conditions

Now, we will apply the initial conditions (IC) to find the specific solution. Using the IC \(y(0) = 3\), we get: \(3 = e^{0}(c_1 \cos(0) + c_2 \sin(0))\) which simplifies to: \(3 = c_1\) Next, we need to find the derivative of the general solution for the second IC. The derivative is: \(y'(t) = e^{2t}(2c_1 \cos(t) - c_1 \sin(t) + 2c_2 \sin(t) + c_2 \cos(t))\) Using the IC \(y'(0) = 5\), we get: \(5 = e^{0}(2(3) \cos(0) - 3 \sin(0) + 2c_2 \sin(0) + c_2 \cos(0))\) which simplifies to: \(5 = 6 + c_2\) Solving for \(c_2\), we get: \(c_2 = -1\)

Step 5: Write the specific solution

Now that we have found the constants \(c_1\) and \(c_2\), we can write the specific solution: \(y(t) = e^{2t}(3 \cos(t) - \sin(t))\) This is the specific solution for the given initial-value problem.

Key Concepts

Complex Roots

In this problem, the characteristic equation yields complex roots. Imagine you solve a quadratic equation, but the numbers don't just fit neatly into the real line. Instead, they take a twist, appearing as complex numbers with both a real part and an imaginary part. Here, we found the roots to be \( r_1 = 2 + i \) and \( r_2 = 2 - i \).

• The real part is '2', which tells us how quickly solutions decay or grow over time.
• The imaginary part is 'i', which introduces oscillations or wave-like repetitions in the solution.

This gives a special structure to the solutions because the imaginary component will influence the behavior of the solution as oscillatory around the real part.

Characteristic Equation

The characteristic equation is the backbone for solving linear differential equations like the one in this exercise. It is derived from the coefficients of the differential equation by substituting a trial solution of the form \( y = e^{rt} \) where \( r \) represents the roots. Here, the original differential equation \( y'' - 4y' + 5y = 0 \) translates into \( r^2 - 4r + 5 = 0 \) for the characteristic equation.

• This quadratic equation essentially defines the properties of solutions we expect, such as stability and oscillatory behavior.
• Solving this equation allows us to determine the nature of the roots—whether they are real or complex.

The solution to these characteristic roots sets the stage for constructing general and specific solutions to the differential equation.

General Solution

The general solution represents a broad spectrum of possible solutions for the differential equation. It encapsulates all the functional forms that satisfy the differential equation for any initial condition. Given complex roots like in our example, it looks like this: \( y(t) = e^{2t}(c_1 \cos(t) + c_2 \sin(t)) \).

• The exponential term \( e^{at} \) accounts for growth or decay based on the real part, '2'.
• The trigonometric terms involve sine and cosine due to the imaginary '1', introducing periodic oscillations.
• Constants \( c_1 \) and \( c_2 \) are undetermined until initial conditions are applied.

This form is key because it combines aspects of exponential growth and oscillation to capture the behavior for any possible initial state.

Specific Solution

By applying the initial conditions (IC) provided, a specific solution is tailored to meet the criteria of the initial-value problem. We use the ICs \( y(0)=3 \) and \( y'(0)=5 \) to pinpoint the exact values of \( c_1 \) and \( c_2 \).

• First, with \( y(0)=3 \), we solve \( 3 = c_1 \), finding \( c_1=3 \).
• Then, using \( y'(0)=5 \), we substitute into the derivative, leading to \( 5 = 6 + c_2 \), and find \( c_2 = -1 \).

The specific solution thus becomes \( y(t) = e^{2t}(3 \cos(t) - \sin(t)) \).This function respects the initial conditions and accurately describes the system's behavior from its particular starting point.