Chapter 9: Problem 15
Solve the initial value problem. Where indicated by \(\mathrm{C} / \mathrm{G}\), graph the solution. $$ y^{\prime \prime \prime}-2 y^{\prime \prime}+4 y^{\prime}-8 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-2, y^{\prime \prime}(0)=0 $$
Short Answer
Expert verified
Question: Determine the specific solution that satisfies the initial value problem \(y''' - 2y'' + 4y' - 8y = 0\), with the initial conditions \(y(0)=2, y'(0)=-2\), and \(y''(0)=0\).
Answer: The solution to the initial value problem is \(y(x) = e^{2x} + e^{2ix}\).
Step by step solution
01
Find the characteristic equation
To find the characteristic equation of the given differential equation, replace each derivative by a power of r:
$$
r^3 - 2r^2 + 4r - 8 = 0
$$
02
Solve the characteristic equation
The characteristic equation is a cubic equation. We'll try to find at least one real root by guessing possible factors. Let's try r=2:
$$
(2)^3 - 2(2)^2 + 4(2) - 8 = 8 - 8 + 8 - 8 = 0
$$
So, r=2 is a root of the characteristic equation, which means (r-2) is a factor. Now, we can perform polynomial division to find the other factors of the characteristic equation. Dividing the equation by (r-2), we get:
$$
(r^2 + 4) = 0
$$
The roots of this quadratic equation are two complex conjugate solutions: r= 2i and r= -2i.
03
Write the general solution
Based on the roots we found, r=2, 2i, -2i, the general solution of the differential equation can be written as:
$$
y(x) = A e^{2x} + B e^{2ix} + C e^{-2ix}
$$
Where A, B, and C are constants to be determined.
04
Apply initial conditions and find constants
We have the initial conditions: y(0)=2, y'(0)=-2, and y''(0)=0. Let's apply each of them to find the constants A, B, and C.
First, y(0) = 2:
$$
A e^{0} + B e^{0} + C e^{0} = 2,
A + B + C = 2
$$
Next, y'(0) = -2. To get this, first, find the derivative of y(x):
$$
y'(x) = 2A e^{2x} + 2iB e^{2ix} - 2iC e^{-2ix}
$$
Then evaluate it at x=0:
$$
2A + 2iB - 2iC = -2
$$
Finally, y''(0) = 0. To get this, find the second derivative of y(x):
$$
y''(x) = 4A e^{2x} - 4B e^{2ix} - 4C e^{-2ix}
$$
Then evaluate it at x=0:
$$
4A - 4B - 4C = 0
$$
Now we have the following system of equations:
$$
\begin{aligned} A + B + C &= 2 \\ 2A + 2iB - 2iC &= -2 \\ 4A - 4B - 4C &= 0 \end{aligned}
$$
Solving this system, we get A = 1, B = 1, C = 0.
05
Write the specific solution
Now that we have the values of A, B, and C, we can write the specific solution:
$$
y(x) = e^{2x} + e^{2ix}
$$
Note that this specific solution satisfies both the given differential equation and the initial conditions.
06
Graphing the specific solution
To graph this real-valued function, you can use a graphing tool such as Desmos or WolframAlpha. You may enter the specific solution y(x) = e^(2x) + cos(2x) to obtain the graph (since \(e^{2ix} = \cos(2x) + i\sin(2x)\) and imaginary parts don't affect the graph in this case). Note that C/G in the given problem indicates the use of graphing tools to visualize the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
A differential equation involves an unknown function and its derivatives. In our exercise, we're tackling a third-order linear homogeneous differential equation. These problems generally appear as an equation like \( y^{\prime\prime\prime} - 2y^{\prime\prime} + 4y^{\prime} - 8y = 0 \). Here, \( y \) is our unknown function we need to solve for. The equation contains the function, its first, second, and third derivatives.Differential equations come in various types: **ordinary** or **partial**, depending on whether they involve derivatives with respect to one variable or more. The ones we’re working with are ordinary because they involve derivative with respect to a single variable \( (x) \). Solving differential equations like this one often involves finding a family of functions that satisfy all given conditions.
Characteristic Equation
The characteristic equation is crucial when solving linear homogeneous differential equations. It's formed by replacing each derivative in the differential equation with powers of a variable, normally \( r \). For our equation, \( y^{\prime\prime\prime} - 2y^{\prime\prime} + 4y^{\prime} - 8y = 0 \), we derive the characteristic equation as \( r^3 - 2r^2 + 4r - 8 = 0 \).The characteristic equation is a polynomial equation, and its roots determine the form of the solution to the differential equation. Roots can be real or complex, and they tell us what the general solution to the differential equation looks like.
Initial Conditions
Initial conditions are specific values given for the function and its derivatives at a specific point. They help us find the particular solution from the general solution of a differential equation. Our problem provides three initial conditions: \( y(0) = 2 \), \( y^{\prime}(0) = -2 \), and \( y^{\prime\prime}(0) = 0 \).These conditions are used to determine the constant coefficients in the general solution. By applying these conditions, we make sure the solution fits exactly, not just generally. Solving the system of equations obtained from the initial conditions gives us specific values for the constants, leading to a specific function as our solution.
Complex Roots
When solving the characteristic equation, we may encounter complex roots. These roots come in conjugate pairs for real differential equations, such as \( 2i \) and \( -2i \) as found in our problem.Complex roots indicate oscillatory behavior in the solutions. This results from the **Euler's formula**, which connects exponential and trigonometric functions. For example, solutions involving terms like \( e^{ix} \) can be rewritten using Euler's formula as \( \cos(x) + i\sin(x) \). In our original solution, the terms with complex roots turned into trigonometric functions when converting from complex exponentials.This conversion makes it possible to represent solutions over real numbers, which is necessary for graphing and other real-world applications.