Chapter 3: Problem 19
Find the solution of the initial value problem $$ y^{\prime \prime}-y=0, \quad y(0)=\frac{5}{4}, \quad y^{\prime}(0)=-\frac{1}{4} $$ $$ \text { Plot the solution for } 0 \leq t \leq 2 \text { and determine its minimum value. } $$
Short Answer
Expert verified
Answer: The minimum value of the function is \(\frac{5}{4}\), which occurs at \(t=0\).
Step by step solution
01
Solve the characteristic equation
To solve the homogeneous equation, first write down its characteristic equation:
$$
r^2 - 1 = 0
$$
To find the roots, factor this equation:
$$
(r - 1)(r + 1) = 0
$$
Now, solve for \(r\), which gives the roots:
$$
r_1 = 1 \quad \text{and} \quad r_2 = -1
$$
02
Write the general solution
Since the roots \(r_1\) and \(r_2\) are distinct real numbers, the general solution of the homogeneous equation takes the following form:
$$
y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}
$$
Plugging in the roots, we get:
$$
y(t) = C_1 e^t + C_2 e^{-t}
$$
03
Apply the initial conditions to find the specific solution
Use the initial conditions to find the constants \(C_1\) and \(C_2\).
First, use the initial condition \(y(0) = \frac{5}{4}\), plug in \(t = 0\):
$$
\frac{5}{4} = C_1 e^0 + C_2 e^0 \implies \frac{5}{4} = C_1 + C_2
$$
Next, differentiate the general solution with respect to \(t\) to find the first derivative \(y'(t)\):
$$
y'(t) = C_1 e^t - C_2 e^{-t}
$$
Use the initial condition \(y'(0) = -\frac{1}{4}\), plug in \(t = 0\):
$$
-\frac{1}{4} = C_1 e^0 - C_2 e^0 \implies -\frac{1}{4} = C_1 - C_2
$$
Now we have two linear equations with two unknowns:
$$
\begin{cases}
C_1 + C_2 = \frac{5}{4} \\
C_1 - C_2 = -\frac{1}{4}
\end{cases}
$$
Solve this system of equations to find \(C_1\) and \(C_2\):
$$
C_1 = 1, \quad C_2 = \frac{1}{4}
$$
04
Write the specific solution
Substitute the values of \(C_1\) and \(C_2\) into the general solution:
$$
y(t) = e^t + \frac{1}{4} e^{-t}
$$
This is the specific solution of the IVP.
05
Plot the solution and find the minimum value
Use mathematical software like Mathematica, Matplotlib, or Desmos to plot the function \(y(t)\) for \(0 \leq t \leq 2\). You will observe that the function has a minimum value for \(t\) in the given interval.
To find the minimum value, differentiate the specific solution with respect to \(t\) again:
$$
y'(t) = e^t - \frac{1}{4} e^{-t}
$$
Set \(y'(t) = 0\)
$$
e^t - \frac{1}{4}e^{-t} = 0
$$
To solve this equation, we can make a substitution: \(u=e^t \implies u^{-1}=e^{-t}\). The equation becomes:
$$
u - \frac{1}{4}u^{-1} = 0
$$
Multiplying by \(u\), we get:
$$
u^2 - \frac{1}{4} = 0 \implies u = \frac{1}{2}
$$
Now, substitute back for \(t\):
$$
e^t = \frac{1}{2} \implies t = \ln{\frac{1}{2}} \approx -0.693
$$
However, the minimum value should be in the interval \(0\leq t \leq 2\). This means there is no critical point in the interval. Therefore, the minimum value will occur at the boundary points.
Evaluate the function at \(t=0\) and \(t=2\):
$$
y(0) = 1 + \frac{1}{4} = \frac{5}{4}
$$
$$
y(2) = e^2 + \frac{1}{4}e^{-2}
$$
So the minimum value of the solution in the interval \(0 \leq t \leq 2\) is indeed \(\frac{5}{4}\), which occurs at \(t=0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
The characteristic equation is a fundamental concept in solving linear homogeneous differential equations. It forms the basis upon which solutions are built. When faced with a second-order homogeneous equation like \(y'' - y = 0\), the first step is to express the problem in terms of its characteristic equation. This process involves replacing each derivative of the function \(y\) with powers of a variable \(r\) (associated with the term \(e^{rt}\) in potential solutions).
For example, for the equation \(y''-y=0\), the characteristic equation becomes \(r^2 - 1 = 0\). By solving this polynomial equation, you determine the roots, which inform the form of the solution to the differential equation. Here, the roots are \(r_1 = 1\) and \(r_2 = -1\), indicating distinct real roots. This process is essential as it provides the crucial building blocks for understanding more complex systems and finding their solutions.
For example, for the equation \(y''-y=0\), the characteristic equation becomes \(r^2 - 1 = 0\). By solving this polynomial equation, you determine the roots, which inform the form of the solution to the differential equation. Here, the roots are \(r_1 = 1\) and \(r_2 = -1\), indicating distinct real roots. This process is essential as it provides the crucial building blocks for understanding more complex systems and finding their solutions.
Initial Value Problem
An Initial Value Problem (IVP) in differential equations refers to a problem where the solution must satisfy not only the differential equation but also specific initial conditions. In the context of the exercise, these initial conditions are given as \(y(0)=\frac{5}{4}\) and \(y'(0)=-\frac{1}{4}\).
These conditions mean that at time \(t = 0\), the function \(y\) has a value of \(\frac{5}{4}\) and its first derivative has a value of \(-\frac{1}{4}\). The purpose of initial conditions is to determine the specific solution of the differential equation from the general solution. Without these conditions, the solution is only general, with constants that can represent any family of solutions.
To solve the IVP, you substitute these conditions into the general solution to solve for the unknown constants, which provides a unique solution that fits the specific scenario described by the initial conditions.
These conditions mean that at time \(t = 0\), the function \(y\) has a value of \(\frac{5}{4}\) and its first derivative has a value of \(-\frac{1}{4}\). The purpose of initial conditions is to determine the specific solution of the differential equation from the general solution. Without these conditions, the solution is only general, with constants that can represent any family of solutions.
To solve the IVP, you substitute these conditions into the general solution to solve for the unknown constants, which provides a unique solution that fits the specific scenario described by the initial conditions.
Homogeneous Equation
A homogeneous differential equation, like \(y'' - y = 0\), is one where the set of outputs for the function yields zero when inputted back into the equation with all possible solutions. This property makes solving such equations more straightforward compared to non-homogeneous ones.
The equation is considered homogeneous because it relates the function \(y\), its derivatives, and possibly higher derivatives, in a way that all terms vanish if the function itself is zero. This allows us to use characteristic equations and guess solutions of the form \(y = e^{rt}\) directly.
Homogeneous equations typically have solutions that are linear combinations of functions, depending on whether the roots of the characteristic equation are real, complex, or repeated. In this case, with distinct roots \(1\) and \(-1\), the general solution is a linear combination of exponentials: \(y(t) = C_1 e^t + C_2 e^{-t}\). This form efficiently leverages the nature of homogeneous equations, offering a powerful tool for solving them.
The equation is considered homogeneous because it relates the function \(y\), its derivatives, and possibly higher derivatives, in a way that all terms vanish if the function itself is zero. This allows us to use characteristic equations and guess solutions of the form \(y = e^{rt}\) directly.
Homogeneous equations typically have solutions that are linear combinations of functions, depending on whether the roots of the characteristic equation are real, complex, or repeated. In this case, with distinct roots \(1\) and \(-1\), the general solution is a linear combination of exponentials: \(y(t) = C_1 e^t + C_2 e^{-t}\). This form efficiently leverages the nature of homogeneous equations, offering a powerful tool for solving them.
General Solution
The general solution of a differential equation encompasses all possible solutions. It includes arbitrary constants that reflect the infinite possible solutions arising from the nature of differential equations. For the homogeneous equation \(y'' - y = 0\), the characteristic equation has distinct real roots \(1\) and \(-1\).
This leads to the general solution \(y(t) = C_1 e^t + C_2 e^{-t}\), where \(C_1\) and \(C_2\) are constants. These constants are not specified in the general solution and instead are determined using initial conditions.
The general solution represents a family of functions that satisfy the differential equation. Each specific solution within this family meets particular constraints or initial conditions pertinent to the problem statement, like those given in an initial value problem. Thus, the general solution provides a crucial framework that encapsulates all potential behaviors, ready to be narrowed by specific external conditions.
This leads to the general solution \(y(t) = C_1 e^t + C_2 e^{-t}\), where \(C_1\) and \(C_2\) are constants. These constants are not specified in the general solution and instead are determined using initial conditions.
The general solution represents a family of functions that satisfy the differential equation. Each specific solution within this family meets particular constraints or initial conditions pertinent to the problem statement, like those given in an initial value problem. Thus, the general solution provides a crucial framework that encapsulates all potential behaviors, ready to be narrowed by specific external conditions.
