Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 44
Solve the given initial-value problem. $$y^{\prime \prime}-4 y=12 e^{2 t}, \quad y(0)=2, \quad y^{\prime}(0)=3$$.
Short Answer
Expert verified
The solution to the initial value problem \(y'' - 4y = 12e^{2t}\), with the initial conditions \(y(0) = 2\) and \(y'(0) = 3\), is given by \(y(t) = \frac{5}{3}e^{2t} + \frac{1}{3}e^{-2t} + 6te^{2t}\).
Step by step solution
01
Find the complementary function
We first consider the related homogeneous equation: \(y'' - 4y = 0\). To find its complementary function, we assume the function has the form y(t) = e^{rt}, where r is a constant. Plugging this into our equation, we get
\((re^{rt})'' - 4(re^{rt}) = 0\).
Simplifying, we get the auxiliary equation \(r^2 - 4 = 0\). Solving for r, we find that r = 2 and r = -2. Thus, the complementary function is given by \(y_c(t) = C_1 e^{2t} + C_2 e^{-2t}\), where C_1 and C_2 are constants.
02
Determine a particular solution for the nonhomogeneous equation
Next, we need to find a particular solution, y_p(t), for the nonhomogeneous equation \(y'' - 4y = 12e^{2t}\). To do this, we guess a particular solution of the form y_p(t) = Qte^{2t}, where Q is a constant we will find. Now we find the first and second derivatives of y_p(t):
\(y_p'(t) = 2Qte^{2t} + Qe^{2t}\)
\(y_p''(t) = 4Qt^2e^{2t} + 4Qte^{2t} + 2Qte^{2t} = 4Qt^2e^{2t} + 6Qte^{2t}\)
Then, we substitute y_p(t) and its derivatives into the nonhomogeneous equation:
\(y_p'' - 4y_p = (4Qt^2e^{2t} + 6Qte^{2t}) - 4(Qte^{2t}) = 4Qt^2e^{2t} + 2Qte^{2t} = 12e^{2t}\)
Comparing terms, we can find the value for Q:
\(2Qt = 12 \Rightarrow Q = 6\)
Thus, the particular solution y_p(t) is given by \(y_p(t) = 6te^{2t}\).
03
Combine complementary function and particular solution
The general solution y(t) is the sum of the complementary function y_c(t) and the particular solution y_p(t):
\(y(t) = y_c(t) + y_p(t) = C_1 e^{2t} + C_2 e^{-2t} + 6te^{2t}\).
04
Apply initial conditions
We'll now apply the initial conditions, y(0) = 2 and y'(0) = 3 to find the constants C_1 and C_2.
First, using y(0) = 2:
y(0) = C_1 e^{0} + C_2 e^{-0} + 6(0)e^{2(0)} = C_1 + C_2 = 2.
Second, using y'(0) = 3, we calculate the derivative of the general solution:
y'(t) = 2C_1e^{2t} - 2C_2e^{-2t} + 12te^{2t} + 6e^{2t}.
Now plugging in t = 0:
y'(0) = 2C_1e^0 - 2C_2e^{-0} + 12(0)e^{2(0)} + 6e^{2(0)} = 2C_1 - 2C_2 + 6 = 3.
Solving the system of equations for C_1 and C_2, we find that \(C_1 = \frac{5}{3}\) and \(C_2 = \frac{1}{3}\).
05
Write the final solution
Substitute the values of C_1 and C_2 into the general solution:
\(y(t) = \frac{5}{3}e^{2t} + \frac{1}{3}e^{-2t} + 6te^{2t}\).
This is the solution to the initial value problem.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial-Value Problem
The initial-value problem involves finding a solution to a differential equation that includes initial conditions to determine any unknown constants. In this exercise, we have the second-order differential equation \( y'' - 4y = 12e^{2t} \) along with the initial conditions \( y(0) = 2 \) and \( y'(0) = 3 \). These initial conditions are necessary because the solution to a second-order differential equation typically involves two arbitrary constants.
- The initial value \( y(0) = 2 \) tells us the value of the function at \( t = 0 \).
- The initial value \( y'(0) = 3 \) provides information about the slope of the solution at \( t = 0 \).
Complementary Function
The complementary function is the part of the general solution to a differential equation that solves the associated homogeneous equation. For the equation \( y'' - 4y = 0 \) (the homogenous part of our original differential equation), we solve to find the complementary function.
By assuming a solution of the form \( y(t) = e^{rt} \), we derive the characteristic equation \( r^2 - 4 = 0 \), which factors into \( (r-2)(r+2) = 0 \). The roots are \( r = 2 \) and \( r = -2 \).
This results in the complementary function:
By assuming a solution of the form \( y(t) = e^{rt} \), we derive the characteristic equation \( r^2 - 4 = 0 \), which factors into \( (r-2)(r+2) = 0 \). The roots are \( r = 2 \) and \( r = -2 \).
This results in the complementary function:
- \( y_c(t) = C_1 e^{2t} + C_2 e^{-2t} \)
Particular Solution
Finding a particular solution involves identifying a specific solution to the nonhomogeneous differential equation, one that accounts for the non-zero right-hand side, \( 12e^{2t} \) in our case. We propose a solution of the form \( y_p(t) = Qte^{2t} \) because the term \( e^{2t} \) aligns with the constant term in the non-homogeneous equation.
Differentiating \( y_p(t) \) provides:
Differentiating \( y_p(t) \) provides:
- \( y_p'(t) = 2Qte^{2t} + Qe^{2t} \)
- \( y_p''(t) = 4Qt^2e^{2t} + 6Qte^{2t} \)
- \( y_p(t) = 6te^{2t} \)
Homogeneous Equation
A homogeneous equation is a differential equation of the form where the right-hand side is zero, such as \( y'' - 4y = 0 \). Solving this is the first step in tackling any non-homogeneous differential equation because it allows us to determine the complementary function. The reason being, any nonhomogeneous differential solution can be expressed as the sum of the complementary function and a particular solution.
The process typically involves:
The process typically involves:
- Setting up and solving the characteristic equation derived from assuming solutions of the form \( e^{rt} \).
- Finding the roots to determine the basic solution structure.
