Using the Complex Exponential

Using the Complex Exponential In the Method of Undetermined coefficients, if the forcing function is an exponential function, then our guess is straightforward. We guess that the particular function is a constant times the exponential, differentiate that twice, and substitute into the DE to find the constant. For example, given y 00 + 2y 0 + y = e2t , the homogeneous part of the solution has r = −1, −1, so that yp0 = 2Ae2t ⇒ yp00 = 4Ae2t Ae2t (4 + 2(2) + 1) = e2t ⇒ A= yp = Ae2t ⇒ and: 1 9 Therefore, yp = 19 e2t In class, we said that if the forcing function was either cos(at) or sin(at), then our ansatz would be: yp = A cos(at) + B sin(at). We then differentiate twice and substitute. However, if we’re comfortable with the complex numbers and the complex exponential, then we can use a shortcut. Here are two related examples. Example 1 Solve y 00 + 2y 0 + y = cos(3t) SOLUTION: Rather than solve this problem, we will solve a related problem by changing the right side of the equation into an exponential function! 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 ) = − 4 3 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 Solve: y 00 + 9y = sin(3t). If we were to do this using sine and cosine, we would have to guess: yp = t(A cos(3t) + B sin(3t)) But wouldn’t it be a bit easier to solve using the complex exponential? In that case, we solve the following, again noting that the homogeneous part of the solution is yh = C1 cos(3t) + C2 sin(3t) We re-write the ODE as: y 00 + 9y = e3it And take the following as our ansatz (multiplied by t). Recall that we’ll actually only need the imaginary part of yp : yp = Ate3it yp0 = Ae3it + 3iAte3it yp00 = 6iAe3it − 9Ate3it Now, yp00 + 9yp = 6iAe3it ⇒ 6iAe3it = e3it . From this, we see that: A= 1 1 =− i 6i 6 As before, we want 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 Example 4 Let’s try another, where we will assume the right side of the equation is the cosine. y 00 + y 0 − 2y = cos(2t) Now, using our trick of embedding this problem into a larger problem: y 00 + y 0 − 2y = cos(2t) + i sin(2t) = e2it 2 The ansatz and its derivatives: yp0 = 2iAe2it yp = Ae2it yp00 = −4Ae2it Putting these into the DE and solve for A: Ae2it (−4 + 2i − 2(1)) = e2it ⇒ A= 1 −6 + 2i Then find the real part of Ae2it :   −6 − 2i 6 2 Real (cos(2t) + i sin(2t)) = − cos(2t) + sin(2t) 40 40 40 I’ve left the fractions unsimplified so that you can see what was done. Example 5 The complex exponential can also be used to help compute integrals involving the sine and cosine. As a simple example:  Z   Z 1 3it 1 e = sin(3t) + C cos(3t) dt = Real e3it dt = Real 3i 3 We can do the same with some harder integral- Namely, an exponential times sine or cosine. We’ll embed this into a larger problem: Z Z Z Z −t −t −t 2it e sin(2t) dt ⇒ e (cos(2t) + i sin(2t)) dt = e e dt = e(−1+2i)t dt And the exponential is easy to integrate. The solution is the imaginary part of: 1 −1 − 2i e2it = (cos(2t) + i sin(2t)) −1 + 2i 5 Therefore, Z 1 2 e−t sin(2t) dt = − cos(2t) − sin(2t) + C 5 5 We’ll take a look at one more example that will be very useful in Section 3.8. Section 3.8 Shortcut Suppose we have the DE: y 00 + py 0 + qy = cos(omegat) where we assume p, q are real numbers. If we use our shortcut assumption, we can solve for the constant A coming from the Method of Undetermined Coefficients: yp = Aeiωt which leaves us with: A= ⇒ Aeiωt (−ω 2 + iωp + q) = eiωt 1 1 = (q − ω 2 ) + iωp α + iβ We will show that, if we write yp = R cos(ωt − δ), then 1 R= |α + iβ| and   β δ = arg(α + βi) = arctan α 3 Proof: Compute the real part:   α − βi α β Real (cos(ωt) + i sin(ωt) = 2 cos(ωt) + 2 sin(ωt) = R cos(ωt − δ) 2 2 2 α +β α +β α + β2 with s R= 1 α2 β2 1 = + =p 2 2 (α2 + β 2 )2 (α2 + β 2 )2 |α + βi| α +β And for the phase angle δ, the sum of squares terms cancel out leaving us with the angle for α + βi, or   β δ = arctan α Using our previous results... Find yp if y 00 + 2y 0 + y = cos(2t) SOLUTION: Let yp = Ae2it and substitute: Ae2it (−4 + 4i + 1) = e2it ⇒ A= 1 −3 + 4i Therefore, the amplitude and phase shift for yp is: R= √ 1 1 = 5 9 + 16 δ = tan−1 (−4/3) + π and yp = R cos(2t − δ). We see that, given a cosine forcing function allows us to very quickly get the “forced response”, yp . Practice: 1. Use the complex exponential to integrate the following: Z (a) e−2t cos(t) dt Z (b) et/2 sin(3t) dt 2. Use the complex exponential to find yp , given: (a) y 00 + 7y = 3 cos(3t) (b) y 00 + y 0 + 3y = 2 sin(2t) (c) y 00 + 2y 0 + y = cos(2t) 3. Use the complex exponential to find the amplitude and phase angle for the forced response: y 00 + 2y 0 + 2y = cos(t) 4 Solutions to the Practice 1. Use the complex exponential to integrate the following: Z (a) e−2t cos(t) dt SOLUTION: We’ll take the real part of: Z Z −2t it e e dt = e(−2+i)t dt = 1 e(−2+i)t −2 + i Expanding this,      −2 − i 2 1 Real e−2t (cos(t) + i sin(t) = e−2t − cos(t) + sin(2t) + C 5 5 5 Z (b) et/2 sin(3t) dt SOLUTION: Same idea as before: Z Z 1 et/2 e3it dt = e( 2 +3i)t dt = 1 2 1 1 2 e( 2 +3i)t = et/2 (cos(3t) + i sin(3t)) 1 + 6i + 3i A little more simplification: 2e t/2   −1 − 6i (cos(3t) + i sin(3t) 37 And we take the imaginary part, so that:   Z 12 2 t/2 t/2 e sin(3t) dt = e − cos(3t) + sin(3t) + C 37 37 2. Use the complex exponential to find yp , given: (a) y 00 + 7y = 3 cos(3t) SOLUTION: The ansatz for y 00 + 7y = 3e3it would be yp = Ae3it (and we’ll keep the real part). Substitute into the DE and factor out Ae3it : Ae3it (−9 + 7) = 3e3it ⇒ A= −3 2 The particular part of the solution is yp = − 32 cos(3t). (b) y 00 + y 0 + 3y = 2 sin(2t) SOLUTION: The ansatz for y 00 + y 0 + 3y = 2 sin(2t) is yp = Ae2it (imaginary part). Substitute: Ae2it (−4 + 2i + 3)e3it = 2e3it ⇒ A= 2 −1 + 2i We take the imaginary part of the expression below (the complex number has been rationalized): −2 − 4i (cos(2t) + i sin(2t)) 5 5 ⇒ 2 2 yp = − cos(2t) − sin(2t) 5 5 (c) y 00 + 2y 0 + y = cos(2t) SOLUTION: Not quite as easy as before, although not bad. We take y 00 + 2y 0 + y = e2it , and yp = Ae2it . Substitute and solve: Ae2it (−4 + 2(2i) + 1) = e2it ⇒ A= 1 −3 + 4i We want the real part of the product: −3 − 4i 1 (cos(2t) + i sin(2t)) = (cos(2t) + i sin(2t)) −3 + 4i 25 3 which is: yp = − 25 cos(2t) + 4 25 sin(2t) 3. Use the complex exponential to find the amplitude and phase angle for the forced response: y 00 + 2y 0 + 2y = cos(t) For this one, we can use our shortcut. Take the ODE to be: y 00 +2y 0 +2y = eit , so yp = Aeit . Substitute into the DE to get: 1 Aeit = (−1 + 2i + 2) = eit ⇒ A = 1 + 2i The amplitude and phase shift are: R= 1 1 = |1 + 2i| 5 6 δ = tan−1 (2)