Question

Consider the initial value problem $$ 4 y^{\prime \prime}+12 y^{\prime}+9 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-4 $$ (a) Solve the initial value problem and plot its solution for \(0 \leq t \leq 5 .\) (b) Determine where the solution has the value zero. (c) Determine the coordinates \(\left(t_{0}, y_{0}\right)\) of the minimum point. (d) Change the second initial condition to \(y^{\prime}(0)=b\) and find the solution as a function of \(b\). Then find the critical value of \(b\) that separates solutions that always remain positive from those that eventually become negative.

Short answer

Question: (a) Solve the given initial value problem and plot its solution for the interval \(0 \leq t \leq 5\). Given: \(4y''(t) + 12y'(t) + 9y(t) = 0\) with \(y(0) = 1\) and \(y'(0) = -4\). (b) Find the values of \(t\) when the solution has a value of zero. (c) Determine the coordinates of the minimum point of the solution. (d) Modify the initial condition: \(y'(0) = b\). Find the solution and analyze the effect of this change on the solution. Determine the critical value of \(b\).

Step-by-step solution

(a) Solution of the Initial Value Problem

(i) Form the characteristic equation for the given linear homogeneous differential equation: $$ 4m^{2} + 12m + 9 = 0 $$ (ii) Solve the characteristic equation for roots \(m_{1}\) and \(m_{2}\): $$ m = \frac{-12 \pm \sqrt{(-12)^{2} - 4(4)(9)}}{8} $$ (iii) Find the general solution in the form: $$ y(t) = C_{1} e^{m_{1}t} + C_{2} e^{m_{2}t} $$ (iv) Apply the initial conditions to find the constants \(C_{1}\) and \(C_{2}\), then plot the solution for \(0 \leq t \leq 5.\)

(b) Determine Where the Solution has the Value Zero

(i) Set the solution \(y(t)\) equal to zero and find the values of \(t\) when \(y(t) = 0.\)

(c) Determine the Minimum Point Coordinates

(i) Find the first derivative of the solution, \(y'(t).\) (ii) Set \(y'(t) = 0\) to find the critical points. (iii) Check the second derivative to determine if the critical point is a minimum: If \(y''(t) > 0,\) then \((t_{0}, y_{0})\) is a minimum point.

(d) Find the Solution for Modified Initial Condition

(i) Replace the initial condition \(y'(0) = -4\) with \(y'(0) = b.\) (ii) Apply the modified initial condition to the general solution and find the new constants \(C_{1}\) and \(C_{2}.\) (iii) Determine the critical value of \(b\) that separates the solutions.

Key concepts

Characteristic Equation

The characteristic equation is a foundational concept in solving linear homogeneous differential equations. It translates the differential equation into an algebraic equation whose roots determine the behavior of the solution. By setting up the characteristic equation for a second-order linear differential equation, we consider the standard form
\( ay'' + by' + cy = 0 \), where \( a, b, \) and \( c \) are coefficients. The corresponding characteristic equation takes the form
\( am^2 + bm + c = 0 \).
Solving this quadratic determines the roots, which can be real and distinct, real and repeated, or complex. The nature of the roots directly affects the form of the general solution of the differential equation. For example, if the roots are real and distinct, the solution would involve exponential functions of the roots. If they are real and repeated, as in the given exercise, the solution involves an exponential function along with a multiplying polynomial term to account for the multiplicity of the root. In the case of complex roots, the solution involves sine and cosine functions that correspond to the imaginary parts of the roots.
Understanding how to form and solve the characteristic equation is crucial in the process of finding the general solution to a homogeneous differential equation.

Homogeneous Differential Equation

A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives, and it is set equal to zero. The term 'homogeneous' signifies that the equation maintains a consistency in degree across all terms, and this property is crucial in allowing us to find solution functions that describe the behavior of the system modeled by the equation. For instance, the given equation from the exercise,
\(4y'' + 12y' + 9y = 0\),
is homogeneous because all terms involve the function \(y\) or its derivatives, and it equals zero.
The importance of identifying an equation as homogeneous lies in the method of solution. Homogeneous differential equations can be solved by finding the characteristic equation, as previously discussed, and using theory that specifically applies to these types of equations, such as the superposition principle. The superposition principle states that if \(y_1(t)\) and \(y_2(t)\) are solutions to a homogeneous linear differential equation, then any linear combination \( C_1y_1(t) + C_2y_2(t) \) is also a solution. This principle greatly simplifies finding general solutions.

Critical Point

In the study of differential equations, critical points, also known as equilibrium points or stationary points, play an important role in understanding the behavior of solutions. A critical point occurs when the derivative of the solution is zero. This point can correspond to a variety of features in the graph of the solution, such as local maximums, local minimums, or points of inflection.
To determine the critical points, we simply set the first derivative of our solution function equal to zero. For our exercise, the first derivative \(y'(t)\) is set equal to zero and we solve for \(t\) to find the critical points. Once we have them, we use the second derivative test to classify the type of each critical point.

• If \(y''(t) > 0\), the point is a local minimum.
• If \(y''(t) < 0\), the point is a local maximum.
• If \(y''(t) = 0\), further analysis is required to classify the point.

In the context of our given problem, we are focused on the minimum points, as they determine where the graph of the solution dips lowest. This is of particular interest in many practical applications, such as physics and engineering, where it might represent the least energy state or the most stable configuration of a system.