Problem 14 - Differential Equations and Linear Algebra

Question

Determine the general solution to the given differential equation. $$y^{\prime \prime}+10 y^{\prime}+25 y=0$$

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime \prime} + 10y^{\prime} + 25y = 0\) is: \[ y(x) = C_1e^{-5x} + C_2xe^{-5x} \]

Step 1: 1. Write down the characteristic equation

In this case, the given differential equation is: \[ y^{\prime \prime} + 10y^{\prime} + 25y = 0 \] The characteristic equation for this differential equation will be of the form: \[ r^2 + 10r + 25 = 0 \]

Step 2: 2. Find the roots of the characteristic equation

Now we will solve the characteristic equation for r. This is a quadratic equation, and its roots can be found using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, a = 1, b = 10, and c = 25. Plugging these values into the formula, we get: \[ r = \frac{-10 \pm \sqrt{10^2 - 4(1)(25)}}{2(1)} \] Simplify the expression: \[ r = \frac{-10 \pm \sqrt{100 - 100}}{2} \] \[ r = \frac{-10 \pm 0}{2} \] Thus, r = -5, which is a repeated root.

Step 3: 3. Write the general solution of the differential equation

Since we have a repeated root r = -5, the general solution of the given differential equation can be written in the form: \[ y(x) = C_1e^{-5x} + C_2xe^{-5x} \] Where \(C_1\) and \(C_2\) are constants that can be determined based on any given initial conditions.

Key Concepts

Characteristic Equation

When you first come across a second-order linear homogeneous differential equation, like the one presented here, you may wonder how to solve it. The characteristic equation is your starting point. This equation comes from a process of transforming the differential equation into an algebraic equation.

How do you create a characteristic equation? For a differential equation of the form \( y'' + by' + cy = 0 \), you assume that solutions might take the form \( y = e^{rx} \). Substitute this assumed form into the original differential equation, and through simplification using derivatives, you end up with a polynomial equation in \( r \), called the characteristic equation.

• Example: For \( y'' + 10y' + 25y = 0 \), the characteristic equation becomes \( r^2 + 10r + 25 = 0 \).

Solving this characteristic equation helps you find the roots, which are crucial for determining the behavior of the solution of the differential equation.

Repeated Roots

Once you've found the characteristic equation, the next step is to solve it to find the roots. These roots play a key role in constructing the general solution. Roots can be unique, repeated, or complex.

• In our example, solving the characteristic equation \( r^2 + 10r + 25 = 0 \) using the quadratic formula, we find \( r = -5 \) and \( r = -5 \).

Repeated roots, like here where both roots are \( -5 \), indicate a special case. With these repeated roots, typical solutions of the form \( e^{rx} \) are not enough. You need to adjust your approach.

For a double root \( r \), the general solution is devised as:

• \( y(x) = C_1e^{rx} + C_2xe^{rx} \)

This ensures that the second part \( C_2xe^{rx} \) still remains a solution of the differential equation, hence accommodating the repeated root nature.

General Solution

Constructing the general solution is the finishing line in solving your differential equation. The general solution combines all possible solutions into one expression. For differential equations with repeated roots, this result has a specific form.

In our case, since both roots of the characteristic equation are \( r = -5 \), the general solution is:

• \( y(x) = C_1e^{-5x} + C_2xe^{-5x} \)

Here, \( C_1 \) and \( C_2 \) are constants. They are determined by initial conditions you'll be given or that you specify ahead of time. If no initial conditions are provided, they remain as arbitrary constants.

Thus, the general solution represents every possible solution for the differential equation, within the context of the given conditions.