Question

Consider the intial value problem $$ 4 y^{\prime \prime}+4 y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=2 $$ (a) Solve the initial value problem and plot the solution. (b) Determine the coordinates \(\left(t_{0}, y_{0}\right)\) of the maximum point. (c) Change the second initial condition to \(y^{\prime}(0)=b>0\) and find the solution as a function of \(b\). (d) Find the coordinates \(\left(t_{M}, y_{M}\right)\) of the maximum point in terms of \(b\). Describe the dependence of \(t_{M}\) and \(y_{M}\) on \(b\) as \(b\) increases.

Short answer

In summary, we have solved the given IVP and found the solution as a function of different initial conditions. The maximum point coordinates were found to depend on the given initial conditions, in this case, b. The behavior of the maximum point coordinates can be analyzed by observing their expressions and plotting their graphs with respect to the variation of the initial conditions.

Step-by-step solution

(a) Solve the IVP and plot the solution:

To solve the IVP, we must first find the general solution of the ODE. We will use the characteristic equation method for this purpose. 1. Write down the corresponding characteristic equation: $$ 4r^2 + 4r + 1 = 0 $$ 2. Solve the characteristic equation: This is a quadratic equation that we can either solve using the quadratic formula or by factoring. In this case, the quadratic equation does not factor nicely, so we use the quadratic formula: $$ r = \frac{-4 \pm \sqrt{4^2 - 4(4)(1)}}{2(4)} $$ 3. Calculate the roots of the characteristic equation: $$ r = \frac{-4 \pm \sqrt{0}}{8} = -\frac{1}{2}\ , \ -\frac{1}{2} $$ Since the roots are real and equal, the general solution is of the form: $$ y(t) = C_1e^{(-1/2)t} + C_2te^{(-1/2)t} $$ 4. Apply the initial conditions to find \(C_1\) and \(C_2\): First, we'll apply \(y(0) = 1\): $$ 1 = C_1e^{0} + C_2(0)e^{-0} \Rightarrow C_1 = 1 $$ Now we'll apply \(y'(0) = 2\). We first need to find the derivative of \(y(t)\): $$ y'(t) = -\frac{1}{2}C_1e^{(-1/2)t} + \left(-\frac{1}{2}C_2te^{(-1/2)t} + C_2e^{(-1/2)t}\right) $$ Now we plug in \(t=0\) and \(y'(0)=2\): $$ 2 = -\frac{1}{2}(1)e^{0} + C_2e^{0} \Rightarrow C_2 = 2+ \frac{1}{2}= \frac{5}{2} $$ Thus, the solution of the IVP is: $$ y(t) = e^{(-1/2)t} + \frac{5}{2}te^{(-1/2)t} $$ Plotting the solution, we obtain a graph of an increasing-decreasing function.

(b) Find the maximum point coordinates:

To find the maximum point coordinates, we need to find when \(y'(t)=0\). 1. Find \(y'(t)\): $$ y'(t) = -\frac{1}{2}e^{(-1/2)t} + \left(-\frac{1}{2}\cdot \frac{5}{2}te^{(-1/2)t} + \frac{5}{2} e^{(-1/2)t} \right) $$ 2. Set \(y'(t)\) to zero and solve for \(t\): $$ -\frac{1}{2}e^{(-1/2)t} -\frac{5}{4}te^{(-1/2)t} + \frac{5}{2} e^{(-1/2)t} = 0 $$ We can simplify the above equation by multiplying through with -2 to make it easier to see the relations between the terms: $$ e^{(-1/2)t}+5te^{(-1/2)t} - 5e^{(-1/2)t} = 0 $$ Now factor out \(e^{(-1/2)t}\): $$ e^{(-1/2)t}(1+5t -5) = 0 $$ Now \(e^{(-1/2)t} \neq 0\) since the exponential function never equals zero. Therefore, we have to solve: $$ 1 + 5t - 5 = 0 \Rightarrow t = \frac{4}{5} $$ Finally, evaluate \(y(t)\) at \(t=\frac{4}{5}\): $$ y\left(\frac{4}{5}\right) = e^{-\frac{4}{10}} + \frac{5}{2}\left(\frac{4}{5}\right)e^{-\frac{4}{10}} \approx 2.0584 $$ Thus, the maximum point coordinates are \(\left( \frac{4}{5}, 2.0584 \right)\).

(c) Change the second initial condition and find the solution as a function of \(b\):

The new IVP becomes: $$ 4 y^{\prime \prime}+4 y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=b > 0 $$ The general solution does not change, and remains: $$ y(t) = C_1e^{(-1/2)t} + C_2te^{(-1/2)t} $$ We need to apply the new initial condition \(y'(0) = b\) and redo step 4 from (a) to find the particular constants \(C_1\) and \(C_2\). $$ b = -\frac{1}{2}e^{0} + C_2e^{0} \Rightarrow C_2= b+ \frac{1}{2} $$ Thus, the solution as a function of \(b > 0\) is: $$ y(t) = e^{(-1/2)t} + (b+\frac{1}{2})te^{(-1/2)t} $$

(d) Find the max point coordinates in terms of \(b\):

Since we already have the derivative \(y'(t)\) and the general solution as a function of \(b\), we need to set \(y'(t)_b\) to zero, solve for \(t\), and plug that back into \(y(t)_b\). 1. Set \(y'(t)_b\) to zero: $$ -\frac{1}{2}e^{(-1/2)t} -\frac{1}{4}(2b+1)te^{(-1/2)t} + (b+\frac{1}{2}) e^{(-1/2)t} = 0 $$ Multiply through with -4: $$ 2e^{(-1/2)t} + (2b+1)te^{(-1/2)t} - 2(2b+1)e^{(-1/2)t} = 0 $$ Factor out \(e^{(-1/2)t}\): $$ e^{(-1/2)t}(2 + (2b+1)t - 2(2b+1) ) = 0 $$ Solve for \(t\): $$ 2 + (2b+1)t - 2(2b+1) = 0 \Rightarrow t = \frac{4b-3}{2b+1} $$ 2. Evaluate \(y(t)\) at \(t = \frac{4b-3}{2b+1}\): $$ y\left(\frac{4b-3}{2b+1}\right) = e^{-\frac{1}{2}\cdot\frac{4b-3}{2b+1}} + \left(b+\frac{1}{2}\right)\left(\frac{4b-3}{2b+1}\right)e^{-\frac{1}{2}\cdot\frac{4b-3}{2b+1}} $$ The maximum point coordinates in terms of \(b\) are \(\left( \frac{4b-3}{2b+1}, e^{-\frac{1}{2}\cdot\frac{4b-3}{2b+1}} + \left(b+\frac{1}{2}\right)\left(\frac{4b-3}{2b+1}\right)e^{-\frac{1}{2}\cdot\frac{4b-3}{2b+1}} \right)\). As \(b\) increases, \(t_M \approx 2\) and \(y_M\) tends to a limit greater than 1. The behavior of \(t_M\) and \(y_M\) can be best understood by observing their graphs.

Key concepts

Understanding Differential Equations

Differential equations are mathematical tools used to describe various phenomena in fields like physics, engineering, biology, and economics. These equations express the relationship between functions and their derivatives, revealing how a particular quantity changes over time or space.

For example, the differential equation from our exercise, \(4y''+4y'+y=0\), is a second-order linear homogeneous differential equation with constant coefficients. The 'second-order' term indicates that the equation includes the second derivative of the function y(t), which relates to acceleration if y(t) represents position. Being 'linear' means that the equation doesn't involve any products or powers of y(t) or its derivatives. 'Homogeneous' implies that there is no stand-alone function of t on the right side of the equation; all terms involve y or its derivatives.

Understanding differential equations is crucial for solving physical problems because these equations often model the laws of nature. In this case, the initial conditions provided additional specific information that pins down a unique solution, allowing us to predict future behavior based on the dynamic at the initial moment.

Characteristic Equation Method

When dealing with linear homogeneous differential equations with constant coefficients, one practical approach to finding a solution is the characteristic equation method. This involves creating an auxiliary equation with constants in place of the function and its derivatives, turning the problem into an algebraic one.

In our exercise, the differential equation \(4y''+4y'+y=0\) is transformed into the characteristic equation \(4r^2 + 4r + 1 = 0\). This equation is then solved for r, which represents the roots of the differential equation's solution. The nature of these roots (real or complex, distinct or repeated) dictates the form of the general solution.

When the roots are real and distinct, the solution is a combination of exponential functions; for real and repeated roots, as in this case, the solution involves both an exponential function and a term that multiplies the exponential by t. The method is powerful because once we understand its mechanics, we can solve a vast array of differential equations systematically.

Maximum Point Coordinates

In calculus, the maximum point on a graph refers to the peak within a particular range, where the function reaches its highest value. Locating this point often involves taking the derivative of the function and finding where this derivative equals zero (indicating a horizontal tangent) and confirming whether it is a local maximum through additional tests.

Our exercise shows the application of this concept. After solving the initial value problem, we differentiate, locate where the derivative is zero, and then calculate the max point coordinates \((t_{0}, y_{0})\) by putting the value of t back into the original function y(t).

When the initial conditions are modified as in parts c and d of the exercise, we encounter a new parameter, b, which affects the solution y(t). As the initial slope \(y'(0)=b\) varies, the solution's shape changes, and so does the location of the maximum point. Analyzing how \(t_{M}\) and \(y_{M}\) depend on b is central to understanding the system's sensitivity to initial conditions and can be especially critical in fields like physics and engineering, where the initial rate of change could represent an initial force or velocity.