Solve the stated initial-value problem. In addition, graph your solution and discuss its long-te

step 1 Welcome to another differential equation problem. In this one, we have y double prime plus 4 y e

Solve the stated initial-value problem. In addition, graph...

Key Concepts

Graphical Representation and Long-Term Behavior

Graphing the solution of a differential equation can provide valuable visual insights into the dynamics of the system, such as oscillations, growth, decay, or stability. Long-term behavior analysis involves understanding the tendencies of the system as time progresses, including whether the oscillations remain bounded, grow without limit, or decay, which is critical for predicting system stability and physical realizability.

General and Particular Solutions

The general solution to a differential equation represents the family of all possible solutions and is typically expressed as a linear combination of independent solutions. The particular solution is obtained by applying the initial conditions to the general solution, thereby pinpointing the specific solution that describes the behavior of the system under the given circumstances.

Characteristic Equation

The characteristic equation is derived by assuming a solution in exponential form when solving constant coefficient linear differential equations. This algebraic equation, whose roots determine the form of the general solution, plays a crucial role in understanding the type of oscillatory or exponential behavior present in the system. The nature of the roots (real, repeated, or complex) dictates the structure of the solution.

Initial Value Problem (IVP)

An initial value problem involves solving a differential equation using given values of the unknown function and its derivatives at a specific point. This provides both a mathematical framework for solving the equation and a unique solution that satisfies the prescribed conditions. IVPs are fundamental in modeling real-world phenomena where the state of the system is known at an initial time, and its future behavior is determined from that state.

Second-Order Linear Homogeneous Differential Equations

These are differential equations involving the second derivative of the unknown function and with all terms linear in the function and its derivatives. In the homogeneous case, the equation is set equal to zero, which often leads to solutions that are combinations of specific functions (such as sine and cosine) that encapsulate the behavior of the system being modeled.

Transcript

00:01 Welcome to another differential equation problem.

00:02 In this one, we have y double prime plus 4 y equals zero.

00:05 We're also given to initial conditions.

00:07 This will help us find the constants of integration in our problem.

00:10 So to solve this differential equation, we're going to use the characteristic equation.

00:15 So we'll have r squared plus 4 equals 0.

00:19 This is simple to solve.

00:20 If we subtract 4 and take the square root, r will be equal to plus or minus 2i.

00:27 Substivision this into the solution for characteristic equations, we'll have y is equal to c, e to the 2i t for its first branch and just negative 2i t for its second branch.

00:48 I'll worry about that later.

00:50 Let's use oilers formula, which is e to the i theta equals cosine of theta plus i sine of theta.

00:57 Solve this.

00:58 If we realize that theta in this is just 2t, then we can simplify this into y1 equals c1 times cosine of 2t plus i sine of 2t plus i sine of 2t.

01:18 Now if you remember we still have this negative 2 i branch.

01:23 So this will just give us a very similar looking equation of y2 equals c2.

01:31 Times cosine of negative 2t plus i sine of negative 2t.

01:41 Now if we want to really analyze this, we have to remember even and odd functions.

01:48 It might have been a while ago, but if we remember that the cosine of t is equal to the cosine of negative t, because the graph of cosine looks like that's our coordinate axis, the graph of cosine looks like this.

02:04 Nope, almost looks like that, where this value on the right side is equal to the same value on the left side.

02:14 Sign, however, is an odd function.

02:17 So the sine of negative t equals negative sign of t.

02:22 So we can pull out that negative sign.

02:25 And we will get minus isign of t, minus isine of 2t.

02:32 Sorry and this cosine of negative 2t can change into cosine of 2t.

02:40 Our total solution for this is just going to be the linear combination of y1 and y2.

02:49 So let's do that now.

02:51 We'll get, let's do it in red.

02:54 So we get y is equal to c1 plus c2 times cosine of 2t.

03:07 And let's take out this parentheses there and the sign will just turn into a c1i plus c2i times sign of 2t i apologize as an error it's be c1i minus c2i like this great if we rename these constants, the c1 plus c2, just to be c1, and the c1i minus c2i just to be c2, and we'll get the proper solution.

03:54 I'm not going to rewrite it.

03:55 I'm just going to say c1 and c2.

03:59 Great.

03:59 So that is our solution to the differential equation...

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