Question

Consider the initial value problem $$ 9 y^{\prime \prime}+12 y^{\prime}+4 y=0, \quad y(0)=a>0, \quad y^{\prime}(0)=-1 $$ (a) Solve the initial value problem. (b) Find the critical value of \(a\) that separates that become negative from those that are always positive.

Short answer

Based on the given solution, find the critical value of 'a' for which the function y(t) is always positive or eventually becomes negative. The critical value of 'a' is given by: $$ a_{crit} = a + \frac{2}{3}a - \frac{3}{2} $$

Step-by-step solution

Formulate the characteristic equation

The given differential equation is: $$ 9y'' + 12y' + 4y = 0 $$ Let's write the characteristic equation associated with this differential equation: $$ 9r^2 + 12r + 4 = 0 $$

Solve the characteristic equation

Now solve the quadratic equation \(9r^2 + 12r + 4 = 0\). This equation can be factored as \(3(3r^2 + 4r + \frac{4}{3}) = 0\), which gives the roots: $$ r_1 = -\frac{2}{3},\quad r_2 = -\frac{2}{3} $$ As both roots are equal, we have a repeated root case.

Write the general solution

When we have a repeated root, the general solution takes the following form: $$ y(t) = (c_1 + c_2t)e^{rt} $$ Substitute the roots into the formula: $$ y(t) = (c_1 + c_2t)e^{-\frac{2}{3}t} $$

Apply initial conditions

We know that \(y(0) = a > 0\) and \(y'(0) = -1\). Let's apply these conditions to find expressions for \(c_1\) and \(c_2\). For the first initial condition, replace \(t\) with \(0\): $$ y(0) = (c_1 + c_2 \cdot 0)e^{-\frac{2}{3}\cdot 0} = c_1 = a $$ Now we need to find \(y'(t)\): $$ y'(t) = (c_2 - \frac{2}{3}(c_1 + c_2t))e^{-\frac{2}{3}t} $$ Apply the second initial condition, replacing \(t\) with \(0\): $$ y'(0) = (c_2 - \frac{2}{3}c_1)e^{-\frac{2}{3}\cdot 0} = -1 $$ We know that \(c_1 = a\), so we get: $$ c_2 - \frac{2}{3}a = -1 $$ Therefore, \(c_2 = \frac{2}{3}a - 1\).

Write the particular solution

Now we found all the constants and can write the particular solution: $$ y(t) = (a + (\frac{2}{3}a - 1)t)e^{-\frac{2}{3}t} $$

Find the critical value of \(a\)

We need to find the boundary value of \(a\) such that \(y(t)\) changes from a positive to a negative function. We need to find a critical point where \(y(t) = 0\). We can rewrite the equation to make it more convenient: $$ y(t) = e^{-\frac{2}{3}t}(a(1 - \frac{2}{3}t)) $$ It's clear that when \(1 - \frac{2}{3}t > 0\), the function is positive. Thus: $$ 1 - \frac{2}{3}t > 0 $$ Solving for \(t\), we get: $$ t < \frac{3}{2} $$ Looking at the changing signs, we see that \(y(t)\) crosses zero and becomes negative when \(t = \frac{3}{2}\). Therefore, the critical value for \(a\) is: $$ a_{crit} = \frac{y(\frac{3}{2})}{e^{-1}} = \frac{(a+\frac{2}{3}a-\frac{3}{2})e^{-1}}{e^{-1}} = a+\frac{2}{3}a-\frac{3}{2} $$ Hence, the critical value of \(a\) is: $$ a_{crit} = a + \frac{2}{3}a - \frac{3}{2} $$

Key concepts

Characteristic Equation

In solving differential equations, identifying the characteristic equation is a vital step. It forms the foundation upon which the behavior of the solutions is understood. The characteristic equation is derived from a linear homogeneous differential equation by assuming a solution of the form e^{rt}, where r is the root of the characteristic equation.

For the given differential equation 9y'' + 12y' + 4y = 0, we postulate the solution to be e^{rt}. This is substituted back into the differential equation to form the characteristic equation 9r^2 + 12r + 4 = 0. Solving this quadratic provides us the key pieces of information needed to construct the general solution to the differential equation.

General Solution of Differential Equation

The general solution of a differential equation represents all possible solutions to the equation, encompassing various scenarios and initial conditions. For a second-order linear homogeneous differential equation with constant coefficients, as shown in our initial value problem, the general solution is a combination of the fundamental set of solutions to the characteristic equation.

If the roots of the characteristic equation are distinct, the general solution is a linear combination of e^{r_1t} and e^{r_2t}. However, in the case of the given problem where we have a repeated root r = -2/3, the general solution takes a modified form: y(t) = (c_1 + c_2t)e^{rt}. This reflects the need to account for the multiplicity of the root when determining the solution to the differential equation.

Applying Initial Conditions

Initial conditions refer to the values of the function and its derivatives at a specific point, usually the beginning of the interval of interest. They are critical for determining the specific solution that satisfies both the differential equation and the given constraints. Applying these conditions helps in finding the exact values of the arbitrary constants c_1 and c_2 in the general solution.

In our problem, the initial conditions y(0) = a and y'(0) = -1 lead us to a set of equations from which c_1 and c_2 can be uniquely determined. In essence, the initial conditions transform the general solution into a particular solution that uniquely fits the scenario at hand.

Critical Value Analysis

Critical value analysis involves finding particular values of parameters within a differential equation's solution that elicit a change in behavior of the system described by the equation. This is crucial in understanding the points of transition where, for instance, a physical quantity might change its state, direction, or stability.

In the given exercise, we're asked to find the critical value of a that separates solutions that start positive and eventually become negative. By examining the particular solution's expression and its behavior over time, we can identify the critical threshold where the sign change occurs. Critical value analysis often involves setting a function or its derivative equal to zero and solving for the parameter in question, enabling us to determine the precise boundary between different behaviors of the system.