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

Euler's Formula

Euler's formula, which states that e^(i?) = cos(?) + i sin(?), provides the foundation for expressing trigonometric functions as complex exponentials. It is crucial in the complexification process as it allows for a concise handling of sine and cosine components in the forcing term when applying the method of undetermined coefficients.

Particular Solution in Differential Equations

A particular solution is a specific solution to a non-homogeneous differential equation that satisfies the entire equation, including the forcing term. The overall solution of the differential equation is then constructed as the sum of the particular solution and the general solution of the associated homogeneous equation.

Method of Undetermined Coefficients

This technique is used to find a particular solution to non-homogeneous linear differential equations with constant coefficients by assuming a form for the particular solution based on the forcing function. Once a suitable trial solution is chosen, its unknown coefficients are determined by substituting the trial solution into the differential equation and equating coefficients.

Complex-Valued Trial Solution

When the forcing function includes trigonometric functions, it is often beneficial to use a complex exponential representation to simplify the algebra. A complex trial solution, such as one involving e^(ax), can be employed, and the real part is extracted at the end to obtain the desired particular solution. This method leverages the ease of differentiating exponential functions over sine and cosine functions.

Transcript

00:01 To start this problem, let's first find the roots of the auxiliary equation here.

00:05 So the roots of the auxiliary equation here.

00:07 So first, our auxiliary equation is going to be this.

00:10 So we get that r is going to be equal to plus or minus i.

00:13 But here, our function f of x, this is going to have the root.

00:18 So this is a is equal to 1.

00:19 This is b is equal to 2.

00:21 So our root is going to be equal to 1 plus 2i like so, or minus 2i.

00:29 So since this is not equal to our roots of the left -hand side, then our trial solution is just going to be simply z -p of x is going to be equal to, then a -0 -e to the 1 plus 2i -x.

00:48 We're going to plug that into this equation here, the complex valued equation, and then we're going to have is equal to 3e to the 1 -2 .e to the 1 .5.

00:59 Plus 2i x like so.

01:02 So now first let's find zp prime and then zp double prime.

01:06 So that's going to be 1 plus 2 i a knot e to the 1 plus 2 i x and then zp double prime of x.

01:16 This is going to be equal to then 1 plus 2 i squared a knot e to the 1 plus 2 i x.

01:24 Let's simplify this here.

01:26 This is going to be equal to if we do 1 1, so it's going to be 1 plus and then 2 times these 2.

01:35 So that's going to be 2i, and then 2i times 2 is going to be 4i.

01:38 And then this squared is going to be negative 4.

01:42 A, not e to the 1 plus 2i x.

01:45 So that's going to become negative 3 plus 4i.

01:50 So i rewrite this as negative 3 plus 4i.

01:56 So now we can plug these 2 into here.

02:02 So we get negative 3 plus 4i, a 0, e to the 1 plus 2i x, plus a knot, e to the 1 plus 2i x, is equal to 3e to the 1 plus 2i x.

02:22 Now we can cancel out all the e to the 2i, 1 plus 2i x, and then we'll also get here, this a not will add with this negative 3 to become negative 3 to become, negative 2, and then plus 4i, a not is equal to 3, is equal to 3 like so...

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