Chapter 2: Problem 21
Solve the initial value problem, Check that your answer satisfies the ODE as well as the initial conditions. (Show the details of your work.) $$y^{\prime \prime}-2 y^{\prime}-3 y=0, y(0)=2, y^{\prime}(0)=14$$
Short Answer
Expert verified
The solution is \( y(t) = 4e^{3t} - 2e^{-t} \), satisfying both the ODE and initial conditions.
Step by step solution
01
Identify the Characteristic Equation
To solve the second-order linear homogeneous differential equation \( y'' - 2y' - 3y = 0 \), we start by assuming a solution of the form \( y = e^{rt} \). Substituting this into the differential equation gives the characteristic equation \( r^2 - 2r - 3 = 0 \).
02
Solve the Characteristic Equation
Solve the characteristic equation \( r^2 - 2r - 3 = 0 \). Factoring, we find \( (r - 3)(r + 1) = 0 \), which implies \( r = 3 \) or \( r = -1 \).
03
Write the General Solution
The roots \( r = 3 \) and \( r = -1 \) lead to the general solution \( y(t) = C_1 e^{3t} + C_2 e^{-t} \), where \( C_1 \) and \( C_2 \) are constants to be determined by initial conditions.
04
Apply Initial Conditions
Use the initial condition \( y(0) = 2 \). Substituting into the general solution, we get \( C_1 e^0 + C_2 e^0 = C_1 + C_2 = 2 \).
05
Differentiate and Apply Second Initial Condition
Find the first derivative: \( y'(t) = 3C_1 e^{3t} - C_2 e^{-t} \). Use the initial condition \( y'(0) = 14 \). Substituting, \( 3C_1 - C_2 = 14 \).
06
Solve System of Equations
We now have two equations: \( C_1 + C_2 = 2 \) and \( 3C_1 - C_2 = 14 \). Solving simultaneously, first solve \( C_1 + C_2 = 2 \) for \( C_2 \): \( C_2 = 2 - C_1 \). Substitute into the second equation: \( 3C_1 - (2 - C_1) = 14 \), or \( 3C_1 - 2 + C_1 = 14 \), simplifying to \( 4C_1 = 16 \). Thus, \( C_1 = 4 \). Substitute back for \( C_2 \): \( C_2 = 2 - 4 = -2 \).
07
Write the Particular Solution
Substituting the constants into the general solution gives the particular solution \( y(t) = 4e^{3t} - 2e^{-t} \).
08
Verify Solution
Check that this satisfies the differential equation by verifying the left-hand side, using our solution in \( y'' - 2y' - 3y \), equals zero. Calculate derivatives: \( y'(t) = 12e^{3t} + 2e^{-t} \) and \( y''(t) = 36e^{3t} - 2e^{-t} \). Substitute into \( y'' - 2y' - 3y \) and confirm it simplifies to zero. Also, verify initial conditions \( y(0) = 2 \) and \( y'(0) = 14 \) hold.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
In the realm of differential equations, the **characteristic equation** plays a pivotal role, especially when dealing with second-order linear homogeneous differential equations. Imagine you encounter an equation like \( y'' - 2y' - 3y = 0 \). To find solutions, mathematicians start by assuming a solution of the form \( y = e^{rt} \). This simple tool, \( e^{rt} \), allows us to transform our differential equation into a polynomial equation known as the characteristic equation.
- In our example, substituting \( y = e^{rt} \) brings forth the characteristic equation \( r^2 - 2r - 3 = 0 \).
- The power of the characteristic equation lies in its ability to reduce the complexity of differential equations to straightforward algebraic problems.
Second-order Linear Differential Equation
Understanding **second-order linear differential equations** is crucial, as they frequently appear in mathematical modeling and real-world problems. Such an equation includes derivatives up to the second order and can generally be expressed in the form \( a y'' + b y' + c y = 0 \). In our exercise, we worked with \( y'' - 2y' - 3y = 0 \).
- These equations are termed 'linear' because the expressions for \( y \), \( y' \), and \( y'' \) are linear in nature—no terms like \( yy' \) or \( (y'')^2 \) appear.
- The 'second-order' designation simply indicates that \( y'' \) is the highest derivative present.
Homogeneous Differential Equation
A **homogeneous differential equation** is marked by a fascinating uniformity. Specifically, a second-order linear homogeneous differential equation like \( y'' - 2y' - 3y = 0 \) equals zero for all terms. This differs from non-homogeneous equations, where a non-zero function appears on the right side, adding complexity.
- Regardless of the functions \( y(t) \) and their respective derivatives, the equation's sum should naturally equal zero.
- The homogeneous property implies solutions can often be combined, using linear combinations of individual solutions, creating a wide range of possible answers to explore.
- Moreover, such solutions must still satisfy given initial conditions to ensure they describe the specific situation modeled.