Addition to 3.5 The Complex Exponential

Addition to 3.5: The Complex Exponential In the Method of Undetermined Coefficients, if the forcing function is an exponential function, g(t) = eat , then our guess is straightforward- yp (t) = Aeat , then we substitute this into the DE and solve for A. For example, given y 00 + 2y 0 + y = e2t , the homogeneous part of the solution has r = −1, −1, so that yh (t) = C1 e−t + C2 te−t and yp = Ae2t , yp0 = 2Ae2t , yp00 = 4Ae2t Substituting yp into the DE, we solve for A (I’ve also factored out the common term): Ae2t (4 + 2(2) + 1) = e2t ⇒ A= 1 9 Therefore, yp = 19 e2t . If the right-hand side is a sine and/or cosine, we can take an ansatz as a sum of the sine and cosine, but it turns out to be convenient to use the complex exponential instead. Here’s an example: Example 1 Solve y 00 + 2y 0 + y = cos(3t). SOLUTION: We already know how to solve the homogeneous problem (yh is given above). We want to focus now on the particular solution, yp (t)- We will embed the given DE into a bigger problem: y 00 + 2y 0 + y = cos(3t) + i sin(3t) = e3it We don’t need the full solution- Because cos(3t) is the real part of the exponential, we will solve the DE and only use the real part as our answer. In doing this, we’ve changed the forcing function from “trig” type to “exponential type”, so our ansatz is now yp = Ae3it . Continuing, treat the forcing function like a regular exponential and solve for the unknown constant A: yp = Ae3it yp0 = 3iAe3it yp00 = −9Ae3it . Substituting into the DE, we get: Ae3it (−9 + 2(3i) + 1) = e3it ⇒ A= 1 1 1 = −8 + 6i 2 −4 + 3i where I’ve factored out the 1/2 to make the computations easier. Our solution is the real part of Ae3it , so we compute that to get our answer. Leaving off the 1/2, 1 −4 − 3i (cos(3t) + i sin(3t)) = (cos(3t) + i sin(3t)) −4 + 3i 25 If we multiply this out, we get:     4 3 3 4 − cos(3t) + sin(3t) + i − cos(3t) − sin(3t) 25 25 25 25 We now have our answer- Note that we did not actually need to compute the imaginary part, we did so in this case to show the complete computation, if you needed it. In our case, we only need the real part, so the particular solution is the following (remember to put the 1/2 back in) yp = Real(Ae3it ) = − 3 4 cos(3t) + sin(3t) 50 50 1 Example 2 Solve: y 00 + 2y 0 + y = sin(3t) In this case, we would use exactly the same technique as before, but we go after the imaginary part of the solution at the end. That is, we: • Replace sin(3t) by e3it • Use yp = Ae3it . • Take the imaginary part of yp as the solution. And we can see what our solution is: yp = − 4 3 cos(3t) − sin(3t) 50 50 Example 3 (We have to multiply by t) Solve: y 00 + 9y = sin(3t). In this case, sin(3t) is part of the homogeneous solution, so our guess needs to take that into account. We re-write the ODE as: y 00 + 9y = e3it And our ansatz will be: yp0 = Ae3it + 3iAte3it yp = Ate3it yp00 = 6iAe3it − 9Ate3it Solving for our constant: yp00 + 9yp = 6iAe3it ⇒ 6iAe3it = e3it . From this, we see that: A= 1 1 =− i 6i 6 As before, we want only the imaginary part of Ae3it , which in this case will be:   t t yp = Imag(Ate3it ) = Imag − i(cos(3t) + i sin(3t) = sin(3t) 6 6 Trigonometry and a Shortcut In the exercises, we will show that, if F (t) = 1 eiωt α + iβ then the real part of F (t) can be written as: Re(F (t)) = R cos(ωt − δ), where R = 1 , |α + iβ| δ = tan−1 (β/α) Similarly, the imaginary part of F (t) can be written as: Im(F (t)) = R cos(ωt − δ), where R = 1 , |α + iβ| δ = tan−1 (−α/β) These formulae are very useful if we want to avoid a lot of tedious computation. Here are some examples. 2 Example Find the particular solution, and express it as R cos(ωt − δ). y 00 + y 0 − 2y = cos(2t) SOLUTION: Take the right side as ei2t , and the ansatz (with derivatives) is: yp0 = 2iAe2it yp = Ae2it yp00 = −4Ae2it Putting these into the DE and solve for A: 1 −6 + 2i Now, recall our formulas- We have our complex number, so find its magnitude and argument: √ √ √ |α + iβ| = 36 + 4 = 40 = 2 10 δ = tan−1 (−1/3) + π Ae2it (−4 + 2i − 2(1)) = e2it ⇒ A= (We added π since (−6, 2) is in Quadrant II). The particular solution is then:
1 √ cos 2t − (tan−1 (−1/3) + π) 2 10 Example Find the particular solution and express it as R cos(ωt − δ). y 00 − 3y 0 + 2y = sin(3t) SOLUTION: Write the ansatz as yp = Ae3it , and substitute it into the DE like last time: Ae3it (−9 − 3(3i) + 2) = e3it From this, | − 7 − 9i| = solution this time). √ 49 + 81 = √ ⇒ A= 1 −7 − 9i 130 and δ = tan−1 (−7/9) (We’re using the imaginary part of the yp = √ 1 cos(3t − δ) 130 Exercises: 1. For each of the following, write the particular solution as yp = R cos(ωt − δ). (a) (b) (c) (d) y 00 + 7y = 3 cos(3t) y 00 + y 0 + 3y = 2 sin(2t) y 00 + 2y 0 + y = cos(2t) y 00 + 2y 0 + 2y = cos(t) 2. Use the complexification technique to find the particular solution to y 00 + y = e−t cos(t). There will be a bit of algebra involved, but not too bad- Write g(t) as e(α+βi)t 3. If F (t) = 1 iωt , α+iβ e then show (by direction computation) that the real part of F can be expressed as Re(F (t)) = 1 cos(ωt − δ), |α + iβ| where δ = arg(α + iβ) And the imaginary part as: Im(F (t)) = 1 cos(ωt − δ), |α + iβ| 3 where δ = arg(−β + iα)