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Magazine Chapter 12: Problem 17
Solve the initial value problem. \(y^{\prime \prime}(t)+4 y(t)=5 e^{-t}\), with \(y(0)=2\) and \(y^{\prime}(0)=1\)
Short Answer
Expert verified
The solution is \(y(t) = \cos(2t) + \sin(2t) + e^{-t}\).
Step by step solution
01
Solve the Homogeneous Equation
Consider the homogeneous equation \(y'' + 4y = 0\). Assume a solution of the form \(y_h = e^{rt}\). This leads to the characteristic equation \(r^2 + 4 = 0\). Solving gives \(r = \pm 2i\), which corresponds to a general solution \(y_h(t) = C_1 \cos(2t) + C_2 \sin(2t)\) (using Euler's formula for complex roots).
02
Solve the Particular Solution
To find a particular solution \(y_p(t)\), use the method of undetermined coefficients. Assume \(y_p(t) = Ae^{-t}\). Then, \(y_p''(t) = A e^{-t}\). Substitute into the differential equation to get \(A e^{-t} + 4A e^{-t} = 5 e^{-t}\), leading to \(5A = 5\). Solve to find \(A = 1\). Thus, \(y_p(t) = e^{-t}\).
03
Form the General Solution
The general solution of the differential equation is the sum of the homogeneous and particular solutions. Therefore, \(y(t) = C_1 \cos(2t) + C_2 \sin(2t) + e^{-t}\).
04
Apply Initial Conditions
Use the initial conditions \(y(0) = 2\) and \(y'(0) = 1\) to find \(C_1\) and \(C_2\). Substituting \(t = 0\) in the general solution, \(y(0) = C_1 + 1 = 2\), so \(C_1 = 1\). Compute \(y'(t)\), giving \(y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t) - e^{-t}\). Substituting \(t = 0\), \(y'(0) = 2C_2 - 1 = 1\), solve for \(C_2\) to get \(C_2 = 1\).
05
Write the Final Solution
Substitute \(C_1 = 1\) and \(C_2 = 1\) into the general solution to get \(y(t) = \cos(2t) + \sin(2t) + e^{-t}\). This is the solution to the initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equations
To solve a differential equation, we often start with what's called the homogeneous equation. In the given exercise, this is expressed as \(y'' + 4y = 0\). A homogeneous equation is characterized by having all terms that involve the unknown function and its derivatives, equating to zero.
- These equations generally provide the basis or structure of solutions, involving functions that are free of any external input, like forcing terms.
- For our equation, we assume a solution form of \(y_h = e^{rt}\), which helps us derive the characteristic equation \(r^2 + 4 = 0\).
- The solution to this equation involves roots \(r = \pm 2i\), indicating complex numbers that lead to sinusoidal functions through Euler's formula.
- Thus, the general solution to the homogeneous equation is \(y_h(t) = C_1 \cos(2t) + C_2 \sin(2t)\). Here, \(C_1\) and \(C_2\) are arbitrary constants determined by further "initial conditions."
Particular Solution
The particular solution addresses the non-homogeneous part of a differential equation. In this exercise, the equation is \(y'' + 4y = 5e^{-t}\). Here, \(5e^{-t}\) is the "forcing term," affecting the behavior of the system. To identify the particular solution, we often use methods like the method of undetermined coefficients.
- Assume a form for the particular solution, typically based on the form of the forcing term. In this case, use \(y_p(t) = Ae^{-t}\).
- Differentiate this assumed solution to determine its derivatives, then substitute back into the original non-homogeneous differential equation.
- Simplify and solve, as seen with \(A e^{-t} + 4A e^{-t} = 5 e^{-t}\), leading to \(5A = 5\).
- Finally, solve for \(A\), giving \(A = 1\). Thus, we find \(y_p(t) = e^{-t}\) as the particular solution.
Initial Conditions
Initial conditions define specific values of the function and possibly its derivatives at a given point (often, \(t = 0\)). They tailor the general solution of a differential equation to fit unique scenarios. In this problem, you're given \(y(0) = 2\) and \(y'(0) = 1\).
- To find the arbitrary constants \(C_1\) and \(C_2\), substitute the initial conditions into the "general solution" obtained from combining homogeneous and particular solutions: \(y(t) = C_1 \cos(2t) + C_2 \sin(2t) + e^{-t}\).
- Substituting \(t = 0\) yields first equation: \(y(0) = C_1 + 1 = 2\), giving \(C_1 = 1\).
- Next, compute \(y'(t)\) and repeat the process: from \(y'(0) = 2C_2 - 1 = 1\), solve for \(C_2\) resulting in \(C_2 = 1\).
