Problem 50 For each of the following differ... [FREE SOLUTION]

Chapter 7: Problem 50

For each of the following differential equations: a. Solve the initial value problem. b. [T] Use a graphing utility to graph the particular solution. $$ y^{\prime \prime}-2 y^{\prime}+10 y=0 \qquad y(0)=1, \quad y^{\prime}(0)=13 $$

Short Answer

Expert verified

The particular solution is $ y(t) = e^t(\cos(3t) + 4\sin(3t)) $.

Step by step solution

Identify the Characteristic Equation

For a differential equation of the form $ y'' - 2y' + 10y = 0 $, start by writing its characteristic equation: $ r^2 - 2r + 10 = 0 $. This is a quadratic equation obtained by replacing $ y'' $ with $ r^2 $, $ y' $ with $ r $, and $ y $ with 1.

Solve the Characteristic Equation

To find the roots of the characteristic equation $ r^2 - 2r + 10 = 0 $, use the quadratic formula: $ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = -2 $, and $ c = 10 $. This results in $ r = 1 \pm 3i $. The roots are complex: $ r_1 = 1 + 3i $ and $ r_2 = 1 - 3i $.

Construct the General Solution

Since the roots are complex, the general solution to the differential equation is $ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) $ where $ \alpha = 1 $ and $ \beta = 3 $. Thus, $ y(t) = e^{t}(C_1 \cos(3t) + C_2 \sin(3t)) $.

Apply Initial Conditions

To find $ C_1 $ and $ C_2 $, use the initial conditions $ y(0) = 1 $ and $ y'(0) = 13 $. First, substituting $ y(0) = 1 $ into $ e^0(C_1\cos(0) + C_2\sin(0)) = 1 $ gives $ C_1 = 1 $. Then, find the derivative $ y'(t) $ and substitute $ t = 0 $ and $ y'(0) = 13 $ to solve for $ C_2 $.

Compute the Derivative and Second Initial Condition

Calculate the derivative: $ y'(t) = e^t(C_1 (3\sin(3t)) + C_2 (3\cos(3t)) + C_1 \cos(3t) - C_2 \sin(3t)) $. Substituting $ y'(0) = 13 $, $ C_1 = 1 $, $ \cos(0) = 1 $, $ \sin(0) = 0 $ yields: $ e^0(0 + 3C_2 + 1) = 13 $. Solving gives $ 3C_2 + 1 = 13 \Rightarrow 3C_2 = 12 $, thus $ C_2 = 4 $.

Write the Particular Solution

The particular solution is $ y(t) = e^t(\cos(3t) + 4\sin(3t)) $, incorporating the initial conditions and roots obtained.

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Key Concepts

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

Initial Value Problem

An initial value problem for a differential equation is a problem where you're provided with the equation itself, along with specific values for the function and its derivative at a certain point.
These values are called initial conditions.
In our example, the differential equation is given by \[ y'' - 2y' + 10y = 0 \]with the initial conditions\[ y(0) = 1, \quad y'(0) = 13 \].The purpose is to find a solution to the differential equation that satisfies these initial conditions.
This ensures the solution is uniquely determined, which is crucial in many practical applications, like physics and engineering.

Characteristic Equation

A characteristic equation arises when solving linear differential equations with constant coefficients.
It transforms a differential equation into an algebraic equation that is easier to solve.
For the given equation \[ y'' - 2y' + 10y = 0, \]the characteristic equation is \[ r^2 - 2r + 10 = 0. \]To convert from a differential equation to the characteristic equation, replace $ y'' $ with $ r^2 $,$ y' $ with $ r $,and $ y $ with 1.This method leverages the fact that solutions of these types of differential equations are often of the exponential form $ e^{rt} $.

Complex Roots

When the characteristic equation has complex roots, it indicates a special form for the solutions involving both exponential and trigonometric functions.
In our example\[ r^2 - 2r + 10 = 0, \]the roots were found using the quadratic formula:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]resulting in complex roots $ r_1 = 1 + 3i $and $ r_2 = 1 - 3i $.These roots show that the solution will include sines and cosines in addition to the exponential function.
Complex roots typically arise when the discriminant $ b^2 - 4ac $ is negative.

General Solution

The general solution of a differential equation with complex roots includes both trigonometric and exponential components.
The form is \[ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)), \]where $ \alpha $ and $ \beta $ are derived from the roots $ r = \alpha \pm \beta i $.
For the given problem, substituting $ \alpha = 1 $ and $ \beta = 3 $ gives:\[ y(t) = e^t(C_1 \cos(3t) + C_2 \sin(3t)). \]Here, $ C_1 $ and $ C_2 $ are constants that are determined by the initial conditions, ensuring the solution fits these given values precisely.

91影视

For each of the following differential equations: a. Solve the initial value problem. b. [T] Use a graphing utility to graph the particular solution. $$ y^{\prime \prime}-2 y^{\prime}+10 y=0 \qquad y(0)=1, \quad y^{\prime}(0)=13 $$

Short Answer

Step by step solution

Identify the Characteristic Equation

Solve the Characteristic Equation

Construct the General Solution

Apply Initial Conditions

Compute the Derivative and Second Initial Condition

Write the Particular Solution

Key Concepts

Initial Value Problem

Characteristic Equation

Complex Roots

General Solution

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