Question

The graph of a solution \(y(t)\) of the differential equation \(4 y^{\prime \prime}+4 y^{\prime}+y=0\) passes through the points \(\left(1, e^{-1 / 2}\right)\) and \((2,0)\). Determine \(y(0)\) and \(y^{\prime}(0)\).

Short answer

In this problem, we solved a second-order linear homogeneous differential equation given by the equation \(4y^{\prime\prime} + 4y^{\prime} + y = 0\). Using the characteristic equation and the quadratic formula, we found that the general solution has the form \(y(t) = C_1e^{-\frac{1}{2}t} + C_2te^{-\frac{1}{2}t}\). Given that the graph passes through the points \((1, e^{-1/2})\) and \((2,0)\), we were able to determine that the particular solution is \(y(t) = 2e^{-\frac{1}{2}t} - te^{-\frac{1}{2}t}\). Finally, we found that \(y(0) = 2\) and \(y^{\prime}(0) = 1\).

Step-by-step solution

Find the characteristic equation

To get the characteristic equation, we replace the given second-order linear homogeneous differential equation \(4 y^{\prime \prime} + 4 y^{\prime} + y = 0\) with \(4r^2 + 4r + 1 = 0\), where \(r\) is a root.

Solve the characteristic equation

Now, we need to solve the quadratic equation: \(4r^2 + 4r + 1 = 0\). We can use the quadratic formula to find the roots: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = 4,\) and \(c = 1\). \(r = \frac{-4 \pm \sqrt{4^2 - 4(4)(1)}}{2(4)} = \frac{-4 \pm \sqrt{16-16}}{8}\) Since the discriminant is 0, we have a repeated root: \(r = \frac{-4}{8} = -\frac{1}{2}\).

Write the general solution

Since we have a repeated root \(r = -\frac{1}{2}\), the general solution to the differential equation is: \(y(t) = C_1e^{-\frac{1}{2}t}+C_2te^{-\frac{1}{2}t}\) where \(C_1\) and \(C_2\) are constants we need to determine.

Use given points to find constants

We are given that the graph passes through the points \((1, e^{-1/2})\) and \((2,0)\). Substitute these values into the general solution and solve for the constants: For the point \((1, e^{-1/2})\): \(e^{-1/2} = C_1e^{-\frac{1}{2}(1)} + C_2(1)e^{-\frac{1}{2}(1)}\) Simplifying: \(1 = C_1 + C_2\) For the point \((2,0)\): \(0 = C_1e^{-\frac{1}{2}(2)} + C_2(2)e^{-\frac{1}{2}(2)}\) Simplifying: \(0 = C_1e^{-1} + 2C_2e^{-1}\) Now we have a system of linear equations: \(C_1 + C_2 = 1\) \(C_1e^{-1}+2C_2e^{-1} = 0\) Solve this system of linear equations. One way to do is to multiply the first equation by \(e^{-1}\) and subtract the second equation from the first: \(C_1e^{-1}+C_2e^{-1}- (C_1e^{-1}+2C_2e^{-1}) = e^{-1} - 0\) Simplifying: \(-C_2e^{-1} = e^{-1}\) Thus, \(C_2 = -1\) Now substitute the value of \(C_2\) into the first equation to find \(C_1\): \(C_1 -1 = 1\) \(C_1 = 2\)

Find the particular solution

Now that we have found the values of \(C_1\) and \(C_2\), we can write the particular solution: \(y(t) = 2e^{-\frac{1}{2}t} - te^{-\frac{1}{2}t}\)

Find \(y(0)\) and \(y^{\prime}(0)\)

To determine \(y(0)\), simply substitute \(t=0\) into the particular solution: \(y(0) = 2e^{0} - (0)e^{0} = 2\) To find \(y^{\prime}(0)\), we need to find the derivative of the particular solution: \(y^{\prime}(t) = -te^{-\frac{1}{2}t} + 2e^{-\frac{1}{2}t} - e^{-\frac{1}{2}t} = -te^{-\frac{1}{2}t} + e^{-\frac{1}{2}t}\) Now, substitute \(t=0\) into the derivative: \(y^{\prime}(0) = -(0)e^{0}+e^{0}= 1\) So the values of \(y(0)\) and \(y^{\prime}(0)\) are \(2\) and \(1\), respectively.

Key concepts

Characteristic Equation

When dealing with second-order linear homogeneous differential equations, the characteristic equation provides a pathway to finding the general solution.
For the differential equation given in the exercise, \(4y'' + 4y' + y = 0\), the characteristic equation is formed by replacing the derivatives with powers of \(r\), such as \(4r^2 + 4r + 1 = 0\). This equation stems from the notion that solutions of linear homogeneous differential equations look similar to exponential functions.
The characteristic equation is a quadratic equation which, upon solving, gives us the roots \(r\). These roots determine the form of the general solution. In our example, the characteristic equation has a repeated root \(r = -\frac{1}{2}\), indicating a specific form for the general solution.

General Solution

The general solution of a differential equation is its most inclusive form, containing arbitrary constants which can be determined if initial conditions are provided.
For a characteristic equation with a repeated root, such as \(r = -\frac{1}{2}\) in this example, the general solution takes on a special form:

• First part: \(C_1e^{rt}\)
• Second part: \(C_2te^{rt}\)

This gives us \(y(t) = C_1e^{-\frac{1}{2}t} + C_2te^{-\frac{1}{2}t}\).
This kind of solution accounts for the repeated influence of the exponential function due to repeated roots. Here, \(C_1\) and \(C_2\) are constants that we can find using initial conditions such as given points from the graph.

Initial Conditions

Initial conditions help us to find the specific solution of a differential equation from its general form. In the exercise, these conditions were provided by the points \((1, e^{-1/2})\) and \((2,0)\) on the graph of \(y(t)\).
By substituting these points into the general solution, we can establish equations that allow us to solve for the constants \(C_1\) and \(C_2\).
Using the point \((1, e^{-1/2})\), we derive the equation \(C_1 + C_2 = 1\). From the point \((2,0)\), another equation \(C_1e^{-1} + 2C_2e^{-1} = 0\) can be formulated.
Such a system of equations gives value to these constants, leading us to a particular solution. These initial condition applications make the broad general solution unique for a specific scenario.

Homogeneous Equations

Differential equations can be classified as homogeneous when all terms depend linearly on the unknown function and its derivatives, like the examples here.
The given equation \(4y'' + 4y' + y = 0\) is homogeneous because it can be expressed in the form where every term involves the function \(y\) or its derivatives, without any separate functions or constants added.
This classification of the differential equation simplifies the solving process since it means there's no additional non-homogeneous term (right-side function) to deal with.

• Helps to focus on finding the general solution just by solving the characteristic equation.
• Allows usage of initial conditions directly to find particular solutions.

It's important for students to understand that the concept of homogeneity makes an equation simpler and often quicker to handle in many cases.