For all problems below, use a complex-valued trial solution to determine a particular solution t

For all problems below, use a complex-valued trial solution...

Key Concepts

Linear Differential Equations with Constant Coefficients

This concept concerns equations where derivatives of the unknown function appear linearly and the coefficients are constants. It is relevant because the differential equation provided has constant coefficients, which means its characteristic equation can be solved using algebraic techniques, and the general solution can be expressed as a sum of homogeneous and particular solutions.

Non-Homogeneous Differential Equations

Non-homogeneous differential equations include a forcing function or external input (in this case, F? cos(?t)) on the right-hand side. Solving such equations involves finding a particular solution that accounts for the effect of this external forcing, in addition to the homogeneous solution that satisfies the associated homogeneous equation.

Complex-Valued Trial Solutions

Using complex-valued trial solutions involves assuming a solution in the form of an exponential function with imaginary components (e.g., e^(i?t)). This approach simplifies the calculus because the exponential function is easier to differentiate and integrate, and it allows the application of Euler's formula to relate complex exponentials to sines and cosines, making it highly effective for solving differential equations involving trigonometric forcing functions.

Method of Undetermined Coefficients

This method involves selecting a form for the particular solution that mirrors the form of the forcing function and then determining the constants (or coefficients) that make the trial solution satisfy the differential equation. In the context of the given problem, a complex trial solution is chosen, and its coefficients are set by substituting back into the differential equation.

Resonance in Differential Equations

Resonance occurs when the frequency of the external forcing function matches the natural frequency of the system (? = ??). This case requires special consideration because the standard form of the trial solution may become linearly dependent on the homogeneous solution, necessitating a modification (typically, multiplication by t) to obtain a valid particular solution. This concept is critical when analyzing the behavior of systems under periodic forcing.

Transcript

00:01 Okay, so to start this problem, we first are going to take a look at the roots of the auxiliary equation here, which is going to be p of r is equal to r squared plus omega -not squared is equal to 0.

00:14 So we're going to have that r is equal to plus or minus i omega, omega -not.

00:21 Now, this part here has the root b is equal to omega.

00:27 So if b is equal to omega, but we need to take a look at the, or compare the roots here.

00:36 So we'll have plus or minus i omega.

00:40 So if omega is not equal to omega -not, so let's do that case first, omega -0 is not equal to omega -0.

00:51 So if this is not omega -0, then our z -p of t is going to be equal to, then, just a knot e to the omega or i omega like so.

01:08 So this is going to have the, our derivative, zp of t, is equal to, and that should be i omega t.

01:18 So we'll have omega i or i omega, and then a knot e to the i omega t, and then zp double prime of t is going to be equal to then negative omega squared a knot e to the i omega t like so let's plug these into our differential equation which is going to solve our plus omega not squared is equal to or is equal to and then we're going to have um on the right hand side this is equal to f not and then then e to the i omega t like so.

02:03 So, oops, sorry.

02:06 And then there should also be a zp here.

02:09 So if we have negative omega squared, so omega squared, a knot, e to the i omega t, and then plus omega not squared, a not, e to the i omega t, is equal to f, not, e to the i omega t, then we have omega not squared minus omega squared a knot is equal to f not.

02:43 So a not is going to be equal to f not over omega not squared minus omega squared.

02:56 So we're going to have our particular, our complex value particular solution zp of t is equal to f not over omega not squared minus omega squared and then times e to the i omega t.

03:13 So e to the i omega t, remember, we're going to replace that actually with cosine omega t plus i sine omega t.

03:22 Since we have the cosine part, we want the real part.

03:25 So that's going to just be this here.

03:27 So our y of p of t is going to be equal to f not or divided by omega not squared minus omens squared, cosine omega -t.

03:42 And this is for omega -not equal to omega -0.

03:48 So that's for that scenario.

03:52 Now we need to take a look at the case where omega is equal to omega -not.

03:57 So for omega equal to omega -not, then that means our particular solution, c -p of t, then it's going to become a -0 -t, e to the i omega t like so.

04:14 So then zp prime t.

04:19 Okay, so let's do that.

04:21 So we are going to have a first times, derivative first time second is going to be that.

04:28 And then plus, now we're going to have i omega t or sorry, i omega times a not t, e to the i omega t...

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