Problem 4 Find the $\gamma_{p}$ and the ... [FREE SOLUTION]

Chapter 16: Problem 4

Find the $\gamma_{p}$ and the $y_{c \prime}$ the general solution, and the definite solution of each of the following: $$y^{\prime \prime}(t)-2 y^{\prime}(t)-10 y=5 ; y(0)=6, y^{\prime}(0)=8 \frac{1}{2}$$

Short Answer

Expert verified

The definite solution is $ y(t) = e^t(6.5 \cos 3t + 2.5 \sin 3t) - 0.5 $.

Step by step solution

Identify the Differential Equation Type

The given equation is a second-order linear non-homogeneous differential equation: $ y''(t) - 2y'(t) - 10y = 5 $. It can be solved by finding the particular solution $ \gamma_p $ and the complementary function $ y_{c'} $.

Find the Complementary Function

Solve the characteristic equation for the homogeneous part $ y'' - 2y' - 10y = 0 $. The characteristic equation is $ r^2 - 2r - 10 = 0 $. Use the quadratic formula: $ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ to find the roots where $ a=1, b=-2, c=-10 $. The roots are $ r = 1 \pm 3i $, indicating complex roots.

Write the Complementary Function

For complex roots $ \alpha \pm \beta i $, the complementary function is $ y_{c'}(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t) $. Here, $ \alpha = 1 $ and $ \beta = 3 $. Thus, $ y_{c'}(t) = e^t(C_1 \cos 3t + C_2 \sin 3t) $.

Find the Particular Solution

The particular solution involves solving for $ \gamma_p $ using a constant function, since the non-homogeneous part is a constant (5). Assume $ \gamma_p = A $, where the derivative is zero, and substitute back into the differential equation: $ -10A = 5 $, giving $ A = -0.5 $. Thus, $ \gamma_p = -0.5 $.

Form the General Solution

Combine the complementary and particular solutions to form the general solution: $ y(t) = y_{c'}(t) + \gamma_p = e^t(C_1 \cos 3t + C_2 \sin 3t) - 0.5 $.

Apply Initial Conditions

Use the initial conditions $ y(0) = 6 $ and $ y'(0) = 8.5 $ to find $ C_1 $ and $ C_2 $. From $ y(0) = e^0(C_1\cdot 1 + C_2\cdot 0) - 0.5 = 6 $, solve to get $ C_1 = 6.5 $. Differentiate $ y(t) $ to apply $ y'(0) = 8.5 $, leading to $ C_2 = 2.5 $.

Write the Definite Solution

The definite solution incorporating initial conditions is $ y(t) = e^t(6.5 \cos 3t + 2.5 \sin 3t) - 0.5 $.

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Key Concepts

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

Second-Order Differential Equation

A second-order differential equation involves derivatives of a function up to the second order. It can generally be expressed in the form $a y'' + b y' + c y = g(t)$, where $a$, $b$, and $c$ are constants and $g(t)$ is a given function of $t$.
In our exercise, the equation is $y''(t) - 2y'(t) - 10y = 5$. This indicates a relationship between the function $y(t)$, its first derivative $y'(t)$, and its second derivative $y''(t)$.
The solution to a second-order differential equation includes finding both the complementary function and a particular solution, which together form the general solution.

Homogeneous and Non-Homogeneous Equations

When dealing with differential equations, classifying them as homogeneous or non-homogeneous is key to solving them.
A

**Homogeneous Equation**: Looks like $ay'' + by' + cy = 0$ where the right-hand side is zero.
**Non-Homogeneous Equation**: Appears as $ay'' + by' + cy = g(t)$, where $g(t)$ is not zero.

In the problem given, $y'' - 2y' - 10y = 5$ is a non-homogeneous equation as the equation equals 5 rather than zero. To solve this, we treat the homogeneous part, $y'' - 2y' - 10y = 0$, separately from the non-homogeneous part and solve them individually.

Complementary Function

For a second-order differential equation like $y''(t) - 2y'(t) - 10y = 0$, solving the homogeneous version yields the complementary function.
The usual approach involves:

Setting up the characteristic equation by replacing $y''$ with $r^2$, $y'$ with $r$, and $y$ with 1.
Solving $r^2 - 2r - 10 = 0$ gives the roots, which are complex: $1 \pm 3i$.
Utilizing these roots, we write the complementary function as $y_{c'}(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t)$, where $\alpha = 1$ and $\beta = 3$.

This results in $y_{c'}(t) = e^t(C_1 \cos 3t + C_2 \sin 3t)$, representing the general solution to the homogeneous equation.
It captures the natural behavior of the system without external influence from the non-homogeneous term.

Particular Solution

The particular solution $\gamma_p$ addresses the non-homogeneous part of the differential equation. Here, the right side of the equation, usually a function like 5 in this exercise, affects the overall solution.
To find $\gamma_p$,

Assume a form that matches the non-homogeneous term. For constants like 5, let $\gamma_p = A$, a constant.
Substitute $\gamma_p$ into the original non-homogeneous equation and solve for $A$.

For our exercise, setting $-10A = 5$ results in $A = -0.5$. Thus, the particular solution is $\gamma_p = -0.5$.

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91影视

Find the \(\gamma_{p}\) and the \(y_{c \prime}\) the general solution, and the definite solution of each of the following: $$y^{\prime \prime}(t)-2 y^{\prime}(t)-10 y=5 ; y(0)=6, y^{\prime}(0)=8 \frac{1}{2}$$