Problem 29 - 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. $$y^{\prime \prime \prime}+3 y^{\prime \prime}-18 y^{\prime}-40 y=0$$

Step-by-Step Solution

Answer

The three linearly independent solutions are \(y_1(x) = e^{-5x}\), \(y_2(x) = e^{2x}\), and \(y_3(x) = e^{-4x}\). The general solution is given by: $$y(x) = C_1 e^{-5x} + C_2 e^{2x} + C_3 e^{-4x}$$ where \(C_1\), \(C_2\), and \(C_3\) are constants.

Step 1: Plug in the trial function \(y(x) = e^{rx}\) into the given differential equation

To start, let's plug our trial function \(y(x) = e^{rx}\) into the given differential equation: $$y^{\prime \prime \prime}+3 y^{\prime \prime}-18 y^{\prime}-40 y=0$$ Taking the derivatives, we get: \(y'(x) = r e^{rx}\) \(y''(x) = r^2 e^{rx}\) \(y'''(x) = r^3 e^{rx}\) Now, plug these derivatives back into the given equation: $$(r^3 e^{rx}) + 3(r^2 e^{rx}) - 18(r e^{rx}) - 40(e^{rx}) = 0$$

Step 2: Factor out \(e^{rx}\)

We can now factor out \(e^{rx}\) from the left-hand side of the equation: $$e^{rx}(r^3 + 3r^2 - 18r - 40) = 0$$

Step 3: Solve the characteristic equation

Since \(e^{rx} \neq 0\), the characteristic equation for this differential equation is given by the polynomial equation: $$r^3 + 3r^2 - 18r - 40 = 0$$ Using either the rational root theorem, factoring, or a numerical root-finding technique (such as Newton's method), we find the three roots of the characteristic equation to be: $$r_1 = -5,~~~r_2 = 2,~~~ r_3 = -4$$ Now that we have found the roots of the characteristic equation, we can determine the independent solutions requested.

Step 4: Write down the linearly independent solutions

The three linearly independent solutions for \(y(x)\) in the form \(y(x) = e^{rx}\) are: $$y_1(x) = e^{-5x}$$ $$y_2(x) = e^{2x}$$ $$y_3(x) = e^{-4x}$$

Step 5: Write the general solution

Now that we have found the three linearly independent solutions, we can write the general solution as a linear combination of these solutions: $$y(x) = C_1 e^{-5x} + C_2 e^{2x} + C_3 e^{-4x}$$ where \(C_1\), \(C_2\), and \(C_3\) are constants.

Key Concepts

Third-Order Differential Equation

When we talk about a third-order differential equation, we're referring to a relationship that involves the third derivatives of a function. In our case, the function is denoted as y(x), and the differential equation is given by: \begin{align*} y^{\text{'''}} + 3y^{\text{''}} - 18y^{\text{'}} - 40y = 0\text{.}\text{\end{align*}}Such an equation can seem daunting at first, but it follows a consistent process of solving that begins with the assumption of a solution of the form \( y(x) = e^{rx} \).Why this form, you may ask? Exponential functions are particularly handy due to their properties when differentiated: an exponential function's derivative is proportional to the function itself, making the calculations more tractable. A key part of solving these equations is the characteristic equation, derived from plugging the assumed solution into the differential equation and simplifying.

Linearly Independent Solutions

Linear independence is an essential concept when dealing with differential equations, which indicates that no solution in a set can be written as a combination of the others. For the third-order differential equation provided, we are looking for three linearly independent solutions. These solutions, of the form \( y(x) = e^{rx} \), are based on the distinct roots of the characteristic equation. In our exercise, we find that the roots are \( r_1 = -5 \), \( r_2 = 2 \), and \( r_3 = -4 \). Each root provides us with an independent solution: \( y_1(x) = e^{-5x} \), \( y_2(x) = e^{2x} \), and \( y_3(x) = e^{-4x} \). The reason they're independent is that none of these exponentials can be expressed as a multiple or combination of the others.To verify the linear independence, one could use techniques like the Wronskian, but in our case, the distinctness of the roots gives us a clear sign of independence. These independent solutions form a solution set that allows for the construction of the general solution to the differential equation.

General Solution of Differential Equation

The general solution to a differential equation is a mix of its particular solutions, which in the case of a linear homogeneous differential equation, involves a linear combination of linearly independent solutions. Our general solution to the given third-order differential equation integrates the three independent solutions with constants that can be determined by initial conditions or boundary values. Mathematically, it is expressed as:\begin{align*} y(x) = C_1 e^{-5x} + C_2 e^{2x} + C_3 e^{-4x}\text{,}\text{\end{align*}}where \( C_1 \), \( C_2 \), and \( C_3 \) are arbitrary real numbers. These constants represent the infinitely many solutions that satisfy the given differential equation. To pin down a specific solution, you would need additional information, such as values of \( y \) and its derivatives at certain points, which are typically given in initial value or boundary value problems.This superposition approach reflects the principle of superposition in linear systems - a fundamental concept in mathematics that allows us to work out complex problems by breaking them down into simpler components.