Problem 3 (More in the next problem set.) ... [FREE SOLUTION]

Chapter 2: Problem 3

(More in the next problem set.) Verify by substitution that the given functions form a basis. Solve the given initial value problem. (Show the details of your work.) $$\begin{array}{l}y^{\prime \prime}+2 y^{\prime}+2 y=0, e^{-x} \cos x, e^{-x} \sin x \\\y(0)=1, y^{\prime}(0)=-1 \end{array}$$

Short Answer

Expert verified

The solution is $ y(x) = e^{-x} \cos x $.

Step by step solution

Verify the Basis Functions

We need to determine if the functions $ e^{-x} \cos x $ and $ e^{-x} \sin x $ are solutions to the differential equation $ y'' + 2y' + 2y = 0 $. First, calculate the derivatives:For $ y_1 = e^{-x} \cos x $:- First derivative: $ y_1' = -e^{-x} \cos x - e^{-x} \sin x $.- Second derivative: $ y_1'' = e^{-x} \cos x + 2e^{-x} \sin x - e^{-x} \cos x = 2e^{-x} \sin x $.Substitute $ y_1 $, $ y_1' $, and $ y_1'' $ into the ODE:\[ y_1'' + 2y_1' + 2y_1 = 2e^{-x} \sin x + 2(-e^{-x} \cos x - e^{-x} \sin x) + 2e^{-x} \cos x = 0 \]For $ y_2 = e^{-x} \sin x $:- First derivative: $ y_2' = -e^{-x} \sin x + e^{-x} \cos x $.- Second derivative: $ y_2'' = e^{-x} \sin x - 2e^{-x} \cos x - e^{-x} \sin x = -2e^{-x} \cos x $.Substitute $ y_2 $, $ y_2' $, and $ y_2'' $ into the ODE:\[ y_2'' + 2y_2' + 2y_2 = -2e^{-x} \cos x + 2(-e^{-x} \sin x + e^{-x} \cos x) + 2e^{-x} \sin x = 0 \]Both functions satisfy the differential equation, confirming they form a basis.

General Solution of the ODE

Given that $ e^{-x} \cos x $ and $ e^{-x} \sin x $ form a basis, the general solution to the differential equation $ y'' + 2y' + 2y = 0 $ is\[ y(x) = C_1 e^{-x} \cos x + C_2 e^{-x} \sin x \]where $ C_1 $ and $ C_2 $ are constants to be determined by initial conditions.

Apply Initial Conditions

Now, use the initial conditions $ y(0) = 1 $ and $ y'(0) = -1 $ to find the constants. At $ x = 0 $:\[ y(0) = C_1 \cdot e^{0} \cos 0 + C_2 \cdot e^{0} \sin 0 = C_1 = 1 \]\[ y'(x) = -C_1 e^{-x} \cos x - C_1 e^{-x} \sin x - C_2 e^{-x} \sin x + C_2 e^{-x} \cos x \]At $ x = 0 $:\[ y'(0) = -C_1 \cdot 1 \cdot 1 - C_2 \cdot 0 + C_2 \cdot 1 = -C_1 + C_2 = -1 \]Substitute $ C_1 = 1 $ into the second equation:\[ -1 + C_2 = -1 \]\[ C_2 = 0 \]

Write Final Solution

With $ C_1 = 1 $ and $ C_2 = 0 $, the solution meeting the initial conditions is:\[ y(x) = e^{-x} \cos x \]

Conclusion and Verification

The function $ y(x) = e^{-x} \cos x $ satisfies the original differential equation and the given initial conditions. The entire solving process confirms that the functions $ e^{-x} \cos x $ and $ e^{-x} \sin x $ form a basis and solve the initial value problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Basis Functions

When solving differential equations, basis functions are a crucial concept. They are a set of independent solutions that help to construct the general solution of a differential equation. For second-order linear differential equations, like the one in our exercise, we typically need two basis functions.

In the given problem, the functions $ e^{-x} \cos x $ and $ e^{-x} \sin x $ are proposed as basis functions. To verify them, we substitute each function into the original differential equation $ y'' + 2y' + 2y = 0 $. Both need to satisfy the equation individually. This means when each expression is placed into the equation, the result should equal zero.

First, compute the first and second derivatives of each function.
Next, substitute these derivatives, along with the original function, back into the differential equation.
Check if the equation simplifies to zero.

These steps confirm that $ e^{-x} \cos x $ and $ e^{-x} \sin x $ are indeed basis functions, as they satisfy the differential equation.

Initial Value Problem

An initial value problem involves finding a specific solution for a differential equation that not only satisfies the equation itself but also meets certain given conditions at a particular point. These conditions are often called initial conditions.

In our exercise, the initial conditions provided were $ y(0) = 1 $ and $ y'(0) = -1 $. These conditions provide values for the function and its derivative at $ x = 0 $. To solve an initial value problem:

Determine the general solution using the verified basis functions.
Use the initial conditions to solve for any constants present in the general solution.

This process pinpoints the specific solution that fulfills both the differential equation and the initial conditions, thus solving the initial value problem.

General Solution

The general solution of a differential equation is a comprehensive solution that encompasses all possible individual solutions. It involves arbitrary constants which can be adjusted to fit specific initial conditions.

For our differential equation $ y'' + 2y' + 2y = 0 $, we determined that the functions $ e^{-x} \cos x $ and $ e^{-x} \sin x $ act as basis functions. Consequently, the general solution is constructed as:

\[ y(x) = C_1 e^{-x} \cos x + C_2 e^{-x} \sin x \]

$ C_1 $ and $ C_2 $ are constants, representing the family of solutions.
These constants are later determined using initial conditions.

By applying the initial conditions from our initial value problem, we found $ C_1 = 1 $ and $ C_2 = 0 $. Therefore, the particular solution $ y(x) = e^{-x} \cos x $ satisfies both the differential equation and initial conditions.

91影视

(More in the next problem set.) Verify by substitution that the given functions form a basis. Solve the given initial value problem. (Show the details of your work.) $$\begin{array}{l}y^{\prime \prime}+2 y^{\prime}+2 y=0, e^{-x} \cos x, e^{-x} \sin x \\\y(0)=1, y^{\prime}(0)=-1 \end{array}$$

Short Answer

Step by step solution

Verify the Basis Functions

General Solution of the ODE

Apply Initial Conditions

Write Final Solution

Conclusion and Verification

Key Concepts

Basis Functions

Initial Value Problem

General Solution

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