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Magazine Chapter 4: Problem 29
\(y^{\prime \prime}-4 y^{\prime}+4 y=e^{2 t} \tan ^{-1} t\)
Short Answer
Expert verified
The general solution is: \[ y(t) = C_1 e^{2t} + C_2 t e^{2t} + y_p(t) \]
Step by step solution
01
Identify the Type of Equation
This is a non-homogeneous linear differential equation because it can be written in the form: \[ y'' - 4y' + 4y = e^{2t} \tan^{-1} t \]
02
Solve the Homogeneous Equation
First, solve the corresponding homogeneous equation: \[ y'' - 4y' + 4y = 0 \]. This can be solved by finding the characteristic equation: \[ r^2 - 4r + 4 = 0 \]
03
Find Roots of the Characteristic Equation
Solve the characteristic equation for roots. This is a quadratic equation which factors as: \[ (r - 2)^2 = 0 \]. Therefore, the roots are: \[ r = 2 \], a double root.
04
Write General Solution of the Homogeneous Equation
Because the roots are repeated, the general solution to the homogeneous equation is: \[ y_h(t) = C_1 e^{2t} + C_2 t e^{2t} \]
05
Solve the Non-Homogeneous Equation Using Variation of Parameters
Use the method of variation of parameters to find a particular solution: \[ y_p = u_1(t) e^{2t} + u_2(t) t e^{2t} \]. First, find the Wronskian \( W \) of the solutions \( e^{2t} \) and \( t e^{2t} \).
06
Compute the Wronskian
Calculate the Wronskian, which is given by: \[ W = e^{2t}(e^{2t} + 2t e^{2t}) - t e^{2t}(2 e^{2t}) = e^{4t}(1 + 2t) - 2t e^{4t} = e^{4t} \]
07
Find Functions u1 and u2
Using the Wronskian, find the functions \( u_1(t) \) and \( u_2(t) \): \[ u_1(t) = - \int \frac{ t e^{2t} \tan^{-1} t}{e^{4t}} dt \] and \[ u_2(t) = \int \frac{ e^{2t} \tan^{-1} t}{e^{4t}} dt \]
08
Solve for u1 and u2 Integrals
Evaluate the integrals for \( u_1(t) \) and \( u_2(t) \). These integrals may require integration by parts: \[ u_1(t) = -\int t e^{-2t} \tan^{-1} t dt \] and \[ u_2(t) = \int e^{-2t} \tan^{-1} t dt \]
09
Form the Particular Solution
Using the functions found, form the particular solution \( y_p(t) \): \[ y_p(t) = u_1(t) e^{2t} + u_2(t) t e^{2t} \]
10
Form the General Solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions: \[ y(t) = y_h(t) + y_p(t) \]. Thus, \[ y(t) = C_1 e^{2t} + C_2 t e^{2t} + y_p(t) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equation
In a differential equation, a homogeneous equation has the form where all terms involve the function or its derivatives, set to zero. For our example, the homogeneous part of the given equation is: \[ y'' - 4y' + 4y = 0 \] To solve it, we ignore the non-homogeneous term on the right-hand side (\( e^{2t} \tan^{-1} t \)). This reduces the equation to a simpler form to handle.
Characteristic Equation
For linear homogeneous second-order differential equations, solving the characteristic equation helps find the general solution. Converting \( y'' - 4y' + 4y = 0 \) to its characteristic form involves substituting \( y = e^{rt} \). This gives the characteristic equation: \[ r^2 - 4r + 4 = 0 \] Solving \( (r-2)^2 = 0 \), we find \( r = 2 \) (a double root). This indicates a repeated root, altering the solution structure.
Variation of Parameters
Variation of parameters is a technique to find a particular solution to non-homogeneous differential equations. For \( y'' - 4y' + 4y = e^{2t} \tan^{-1} t \), we set \( y_p = u_1(t) e^{2t} + u_2(t) t e^{2t} \). Determining \( u_1(t) \) and \( u_2(t) \) involves integrals influenced by the Wronskian and the non-homogeneous term \( e^{2t} \tan^{-1} t \).
Wronskian
The Wronskian, \( W \), of two functions checks their linear independence and is pivotal in variation of parameters. For our solutions, \( e^{2t} \) and \( t e^{2t} \), the Wronskian is: \[ W = e^{2t} (e^{2t} + 2t e^{2t}) - t e^{2t} (2 e^{2t}) = e^{4t}(1 + 2t) - 2t e^{4t} = e^{4t} \] This result helps us form the equations to find \( u_1(t) \) and \( u_2(t) \).
General Solution
Combining the homogeneous and particular solutions provides the general solution for the differential equation. Using our example: \[ y(t) = C_1 e^{2t} + C_2 t e^{2t} + y_p(t) \] Here, \( C_1 e^{2t} + C_2 t e^{2t} \) solves the homogeneous part, and \( y_p(t) \) incorporates the non-homogeneous term influenced by \( e^{2t} \tan^{-1} t \). The constants \( C_1 \) and \( C_2 \) are determined by initial conditions if given, finalizing the solution.
