91Ó°ÊÓ

A differential equation involves a function and its derivatives. These types of equations describe various phenomena such as motion, heat, and more. In our exercise, the differential equation is: \[ y'' + 4y' + 3y = 0 \] This is a **second-order linear homogeneous differential equation** because it involves the second derivative of \( y \), and it does not contain any terms that are just functions of \( t \) with no \( y \). This equation will be solved to find \( y(t) \), the unknown function.

To solve the differential equation, we first convert it into its characteristic equation. By assuming a solution of the form \( y = e^{rt} \), we substitute \( y \) and its derivatives back into the original equation: \[ y = e^{rt},\text{ } y' = re^{rt}, \text{ } y'' = r^2e^{rt} \] Substituting these into the differential equation gives us: \[ r^2 e^{rt} + 4r e^{rt} + 3 e^{rt} = 0 \] Factor out \( e^{rt} \) (which is never zero): \[ r^2 + 4r + 3 = 0 \] Now we solve this quadratic equation to find the roots, \( r = -1 \) and \( r = -3 \). These roots are crucial as they help us find the general solution to the differential equation.

Local maxima and minima in a function are the points where the first derivative, \( y'(t) \), is zero. To detect these points in our problem, first we need the first derivative of our general solution: \[ y(t) = c_1 e^{-t} + c_2 e^{-3t} \] The first derivative is: \[ y'(t) = -c_1 e^{-t} - 3c_2 e^{-3t} \] Setting \( y'(t) = 0 \) gives points where the slope is zero. These potential points of maxima or minima occur when: \[ -\frac{3a + b}{2} e^{-t} - \frac{3(a + b)}{2} e^{-3t} = 0 \] Solving this, we find: \[ t = \text{ln}(3) \]. Therefore, the conditions for **no** local maxima or minima is when \( b = 3a \). To get exactly one local maximum, we need \( a > 3b \). For exactly one local minimum, \( b > 3a \), ensuring the function has desired properties in intervals.

Initial conditions are the values given at a specific point, usually where \( t = 0 \). For our problem, the initial conditions are given as: \[ y(0) = a, \text{ } y'(0) = b \] Applying these conditions, we use them to find the constants \( c_1 \) and \( c_2 \) in our general solution. Starting with: \[ y(0) = c_1 + c_2 = a \] And then, taking the derivative: \[ y'(0) = -c_1 - 3c_2 = b \] Solving the system of equations: \[ c_1 + c_2 = a \] \[ -c_1 - 3c_2 = b \] yields \( c_2 = -\frac{a + b}{2} \) and \( c_1 = \frac{3a + b}{2} \). These constants help to fully determine the unique solution to our initial value problem.

The general solution to the differential equation comes from the roots of the characteristic equation. With \( r = -1 \) and \( r = -3 \), the solution will be: \[ y(t) = c_1 e^{-t} + c_2 e^{-3t} \] Incorporating initial conditions \( y(0)=a \) and \( y'(0)=b \), we found the constants: \[ c_1 = \frac{3a + b}{2}, \text{ } c_2 = -\frac{a + b}{2} \] Thus, the complete solution: \[ y(t) = \frac{3a + b}{2} e^{-t} - \frac{a + b}{2} e^{-3t} \] This solution meets the conditions of the initial value problem while appropriately addressing the characteristics pointed out in the problem, like the presence or absence of local maxima and minima.