Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 8
Find the general solution to the differential equation. $$y^{\prime \prime}-4 y^{\prime}+3 y=e^{3 t}$$
Short Answer
Expert verified
The general solution is \(y(t) = C_1 e^{3t} + C_2 e^{t} + \frac{1}{2} te^{3t}.\)
Step by step solution
01
Identify the Type of Differential Equation
The given equation is a second-order non-homogeneous linear differential equation with constant coefficients: \[y'' - 4y' + 3y = e^{3t}.\]
02
Solve the Homogeneous Equation
First, solve the associated homogeneous equation:\[y'' - 4y' + 3y = 0.\]The characteristic equation is:\[r^2 - 4r + 3 = 0.\]Factoring gives:\[(r - 3)(r - 1) = 0.\]The solutions are \(r = 3\) and \(r = 1\). The general solution of the homogeneous equation is:\[y_h(t) = C_1 e^{3t} + C_2 e^{t},\]where \(C_1\) and \(C_2\) are constants.
03
Find a Particular Solution
To find a particular solution \(y_p(t)\) to the non-homogeneous equation, consider the form similar to the non-homogeneous part, which is \(e^{3t}\). Since \(e^{3t}\) is already a part of the homogeneous solution, modify by multiplying \(e^{3t}\) by \(t\):\[y_p(t) = At e^{3t}.\]Compute the derivatives:\[y_p'(t) = Ae^{3t} + 3Ate^{3t},\]\[y_p''(t) = 6Ae^{3t} + 3Ate^{3t}.\]Substitute into the differential equation:\[(6Ae^{3t} + 3Ate^{3t}) - 4(Ae^{3t} + 3Ate^{3t}) + 3Ate^{3t} = e^{3t}.\]Simplify and solve for \(A\):\[(6A - 4A)e^{3t} = e^{3t}.\]\[2A = 1, \quad A = \frac{1}{2}.\] Thus, \(y_p(t) = \frac{1}{2}te^{3t}.\)
04
Combine General Solutions
The general solution \(y(t)\) of the non-homogeneous equation is the sum of the homogeneous solution \(y_h(t)\) and the particular solution \(y_p(t)\):\[y(t) = C_1 e^{3t} + C_2 e^{t} + \frac{1}{2} te^{3t}.\]
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.
Non-homogeneous Equations
Non-homogeneous differential equations are a type of differential equation that involves a non-zero term on the right-hand side. In our given example, the equation is \( y'' - 4y' + 3y = e^{3t} \). The term \( e^{3t} \) makes it non-homogeneous.
Non-homogeneous equations typically take the form of \( L[y] = f(t) \), where \( L \) is a linear differential operator, and \( f(t) \) is a function of \( t \), different from zero.
Non-homogeneous equations typically take the form of \( L[y] = f(t) \), where \( L \) is a linear differential operator, and \( f(t) \) is a function of \( t \), different from zero.
- The presence of \( f(t) \) distinguishes non-homogeneous equations from homogeneous equations, where \( f(t) = 0 \).
- Solving these equations involves finding both the particular solution and the homogeneous solution, and then combining them to obtain the general solution.
Characteristic Equations
The characteristic equation is a crucial tool in solving homogeneous linear differential equations. It emerges from the homogeneous counterpart of the given differential equation. For the homogeneous equation \( y'' - 4y' + 3y = 0 \), the characteristic equation is derived by assuming solutions of the form \( e^{rt} \), which leads us to substitute \( y = e^{rt} \) into the equation.
This results in a polynomial equation, termed the characteristic equation: \( r^2 - 4r + 3 = 0 \). By solving this quadratic, we find the roots \( r = 3 \) and \( r = 1 \).
This results in a polynomial equation, termed the characteristic equation: \( r^2 - 4r + 3 = 0 \). By solving this quadratic, we find the roots \( r = 3 \) and \( r = 1 \).
- The solutions to this characteristic equation, known as roots, are fundamental in constructing the homogeneous solution.
- The nature of roots (distinct, repeated, or complex) influences the form of the solution.
Particular Solution
Finding a particular solution involves determining a specific solution to the non-homogeneous equation. For our particular equation \( y'' - 4y' + 3y = e^{3t} \), we initially guessed the solution would resemble \( e^{3t} \), due to the format of the non-homogeneous term.
However, since \( e^{3t} \) is part of the homogeneous solution, we multiply it by \( t \) to avoid duplication. Therefore, we assume a particular solution of the form \( y_p(t) = At e^{3t} \).
However, since \( e^{3t} \) is part of the homogeneous solution, we multiply it by \( t \) to avoid duplication. Therefore, we assume a particular solution of the form \( y_p(t) = At e^{3t} \).
- Compute its derivatives: \( y_p'(t) = Ae^{3t} + 3At e^{3t} \) and \( y_p''(t) = 6Ae^{3t} + 3At e^{3t} \).
- Substitute these back into the differential equation and solve for \( A \).
- This method ensures the particular solution satisfies the non-homogeneous equation.
Homogeneous Solution
A homogeneous solution is a part of the overall solution that satisfies the homogeneous differential equation \( y'' - 4y' + 3y = 0 \). This is derived by solving the characteristic equation, which provides us the form of the solution.
The resulting homogeneous solution for this differential equation is \( y_h(t) = C_1 e^{3t} + C_2 e^{t} \), where \( C_1 \) and \( C_2 \) are constants determined through initial conditions.
The resulting homogeneous solution for this differential equation is \( y_h(t) = C_1 e^{3t} + C_2 e^{t} \), where \( C_1 \) and \( C_2 \) are constants determined through initial conditions.
- Homogeneous solutions are based solely on the differential operator, disregarding any non-homogeneous terms.
- These solutions form the complementary component of the general solution.
