Problem 24 - Differential Equations and Linear Algebra

Question

Determine two linearly independent solutions to the given differential equation of the form \(y(x)=e^{r x},\) and thereby determine the general solution to the differential equation. $$y^{\prime \prime}+7 y^{\prime}+10 y=0$$

Step-by-Step Solution

Answer

The characteristic equation of the given differential equation is \[(r^2) + 7(r) + 10 = 0\], and its solutions are \(r_1=-2\) and \(r_2=-5\). Thus, the two linearly independent solutions are \(y_1(x) = e^{-2x}\) and \(y_2(x) = e^{-5x}\). The general solution is given by: \[y(x) = C_1 e^{-2x} + C_2 e^{-5x}\] where \(C_1\) and \(C_2\) are arbitrary constants.

Step 1: Find the Characteristic Equation

To find the characteristic equation of the given differential equation, we replace \(y^{\prime \prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1. This gives us the characteristic equation: \[(r^2) + 7(r) + 10 = 0\]

Step 2: Solve the Characteristic Equation

We have a quadratic equation, which we can solve for r: \[(r^2)+7r+10 = 0\] Factoring the equation, we get \[(r+2)(r+5)=0\] This gives us two distinct real roots, which are \(r_1=-2\) and \(r_2=-5\).

Step 3: Determine the Linearly Independent Solutions

Using the values of r we obtained in the previous step, we can now determine two linearly-independent solutions to the given differential equation. Recall that the solutions are of the form \(y(x)=e^{r x}\). Plug in the values of r to obtain the linearly independent solutions: - For \(r_1=-2\) \(y_1(x) = e^{-2x}\) - For \(r_2=-5\) \(y_2(x) = e^{-5x}\) So, \(y_1(x)=e^{-2x}\) and \(y_2(x)=e^{-5x}\) are two linearly independent solutions to the given differential equation.

Step 4: Determine the General Solution

The general solution to the given differential equation can be expressed as a linear combination of the two linearly independent solutions we found in the previous step: \[y(x) = C_1 e^{-2x} + C_2 e^{-5x}\] where \(C_1\) and \(C_2\) are arbitrary constants. The general solution to the given differential equation is: \[y(x) = C_1 e^{-2x} + C_2 e^{-5x}\]

Key Concepts

Characteristic Equation

When working with differential equations, particularly second-order linear equations like the one in the exercise, finding the characteristic equation is an essential step. The characteristic equation helps transform a differential equation into an algebraic one, making it easier to solve. In the example given, the differential equation is
\[ y'' + 7y' + 10y = 0 \]
To form the characteristic equation, replace derivatives with powers of a proposed variable, typically denoted as \( r \):

• \( y'' \) becomes \( r^2 \)
• \( y' \) becomes \( r \)
• \( y \) becomes 1

Putting it all together, we develop the characteristic equation:
\[ r^2 + 7r + 10 = 0 \]
This is a quadratic equation, and solving it will provide the roots that are crucial for finding solutions to the original differential equation.

General Solution

Once we have found the roots of the characteristic equation, we can move on to determining the general solution of the differential equation. The general solution is a combination of all possible solutions, capturing the behavior of the differential equation comprehensively.
The characteristic equation provided us with roots \( r_1 = -2 \) and \( r_2 = -5 \). Each root corresponds to a distinct exponential function solution of the form \( y(x) = e^{rx} \).
For our roots:

• Root \( r_1 = -2 \) gives \( y_1(x) = e^{-2x} \)
• Root \( r_2 = -5 \) gives \( y_2(x) = e^{-5x} \)

The general solution combines these linearly independent solutions:
\[ y(x) = C_1 e^{-2x} + C_2 e^{-5x} \]
Here, \( C_1 \) and \( C_2 \) are constants that can be determined if initial conditions are provided.

Linearly Independent Solutions

In the context of a differential equation, linearly independent solutions provide us with sufficient distinct solutions needed to form a general solution. Linear independence means that one solution cannot be expressed as a scalar multiple of the other.
For the differential equation presented:

• The solutions \( y_1(x) = e^{-2x} \) and \( y_2(x) = e^{-5x} \) are linearly independent.
• This is because there is no constant \( C \) such that \( e^{-2x} = C \, e^{-5x} \) for all \( x \).

These distinct solutions allow us to express the general solution as a combination of them, which is crucial. Without linear independence, the general solution would lack completeness, failing to capture all possible scenarios described by the differential equation. This is why finding multiple linearly independent solutions ensures the general solution encompasses every possible outcome.