Problem 28 - Differential Equations and Linear Algebra

Question

Determine three 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. $$y^{\prime \prime \prime}+3 y^{\prime \prime}-4 y^{\prime}-12 y=0$$

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime\prime\prime} + 3y^{\prime\prime} - 4y^{\prime}-12y = 0\) is: \[y(x) = C_1 e^{-6x} + C_2 e^{-x} + C_3 e^{2x}\] where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants, and the three linearly independent solutions are \(y_1(x) = e^{-6x}\), \(y_2(x) = e^{-x}\), and \(y_3(x) = e^{2x}\).

Step 1: Substitute the given form and find the characteristic equation

Let's substitute \(y(x) = e^{rx}\) into the given equation: \((re^{rx})'''+3( re^{rx})''-4( re^{rx})'-12( re^{rx})=0\) Now, by using the exponent property and considering that \(e^{rx} \ne 0\) for any real value of \(r\) and \(x\), we can rewrite the equation as follows: \((r^3 + 3r^2 - 4r - 12)e^{rx}=0\) Since \(e^{rx}\) is never zero, we can divide both sides by it, and then we are left with the characteristic equation: \[r^3 + 3r^2 - 4r - 12 = 0\]

Step 2: Solve the characteristic equation

Now we need to find the roots of the characteristic equation. This step may require some guesswork or using polynomial factoring techniques. In this case, notice that \(r=-1\) is a root. \[((r+1)(r^2 + 2r - 12)) = 0\] Now factor the quadratic term, and we get: \[((r+1)(r - 2)(r + 6))=0\] So, the roots are \(r_1 = -6\), \(r_2 = -1\), and \(r_3 = 2\).

Step 3: Find the three linearly independent solutions

Now we can write the three linearly independent solutions using the roots we found: \(y_1(x) = e^{-6x}\) \(y_2(x) = e^{-x}\) \(y_3(x) = e^{2x}\)

Step 4: Determine the general solution

Finally, the general solution of the given differential equation is a linear combination of the three linearly independent solutions we found. Thus, the general solution is: \[y(x) = C_1 e^{-6x} + C_2 e^{-x} + C_3 e^{2x}\] where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Key Concepts

Characteristic Equation

To solve differential equations of the form given in the exercise, we begin by finding the characteristic equation. The characteristic equation derives from substituting a solution of exponential form, generally expressed as \(y(x) = e^{rx}\), into the original differential equation. By making this substitution and simplifying, we equate to zero any term involving \(e^{rx}\) since \(e^{rx}\) itself is never zero. Take the equation \(y''' + 3y'' - 4y' - 12y = 0\) for example.
By substituting \(y(x) = e^{rx}\), each derivative of \(y\) changes to a power of \(r\), resulting from the repeated application of the chain rule. This transforms the differential equation into a polynomial equation of \(r\). In our case, it becomes \(r^3 + 3r^2 - 4r - 12 = 0\).

• This polynomial, known as the characteristic equation, is key to finding solutions.
• The roots of this characteristic equation will determine the form of the solutions to the differential equation.

Finding these roots might require some trial for integer solutions or polynomial solving techniques. These roots represent the exponents in the exponential solutions of the equation.

Linearly Independent Solutions

Once the characteristic equation is solved, the next step is to find linearly independent solutions. Linear independence in the context of solutions refers to the fact that one solution cannot be expressed as a constant multiplication of another. Each independent solution adds a unique component to the general solution.
For our equation, we found three roots: \(r_1 = -6\), \(r_2 = -1\), and \(r_3 = 2\). These lead to three distinct solutions based on substituting back into the exponential form \(y(x) = e^{rx}\).

• Each solution constructed by these roots is \(y_1(x) = e^{-6x}\), \(y_2(x) = e^{-x}\), and \(y_3(x) = e^{2x}\).
• Since each involves a distinct root, these solutions are linearly independent.

Linear independence ensures that these solutions contribute completely and uniquely to any potential solution space, which is vital when constructing the general solution.

General Solution

The general solution of a differential equation encompasses all possible solutions and is composed of linear combinations of the identified linearly independent solutions. This holds true particularly for linear homogeneous differential equations like the one we're working with.
Once we have the independent solutions, the general solution is formed by their linear combination. We introduce arbitrary constants to represent potential scales and shifts from specific solutions that satisfy initial or boundary conditions.
In our example, the general solution is expressed as:
\[y(x) = C_1 e^{-6x} + C_2 e^{-x} + C_3 e^{2x}\]
where \(C_1, C_2,\) and \(C_3\) are arbitrary constants. They adjust the contribution of each solution to meet specific conditions.

• The general solution fully captures the behavior of the differential equation for any initial conditions.
• The combination of different exponentials and arbitrary constants allows flexibility and adaptation to many scenarios often seen in applications.

This concept of a general solution is crucial as it presents the most comprehensive form of solutions for the differential equation, encompassing all particular solutions.