Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The differential equation MX double prime plus CX prime plus KX equaling F of T is an equation that governs the one-dimensional motion of a mass M that is attached to a spring with constant K and a dash pot with constant C and is also acted by an external force F of T. Machines with rotating components commonly involve mass spring systems or their equivalents in which the external force is simple harmonic. So F of T would equal F naughty cosine Omega T or F of T equal F naughty sine Omega T where the constant F naughty is the amplitude of the periodic force and Omega is at circular frequency. If we think about undamped forced oscillation, that's when C = 0. So that equation then turns into MX double prime plus KX equaling F naughty cosine Omega T. So if we solved for our R, we got Mr. squared plus K equaling 0. So our R would be positive negative sqrt K / m * I. So our complementary function would be XCX sub C equaling C1 cosine the square root of K divided by Mt plus C2 sine the square root of K divided by Mt. Now if we let Omega naughty equal the square root of KM, that would then simplify into XC equaling C1 cosine Omega naughty t + C two sine Omega naughty T. Now looking at the particular or the trial solution, we'd have X of P equaling a cosine Omega t + b sine Omega T, taking the first derivative and the second derivative and then substituting it back into the original equation. So we can solve for our A and our B. We can see that negative AM Omega squared plus AK equal F not OR a is just F not divided by K -, m Omega squared. Now, if we thought about taking everything and dividing by M, we'd get F not divided by M over K / m minus Omega squared. But we know that K / m is really Omega naughty squared, so a new equivalent equation would be F naughty divided by M divided by Omega naughty squared minus Omega squared. So our X, our actual equation is our X complement plus our X trial. So our X of T equals XC plus XP. Now if we look at damped forced oscillation, damped forced oscillation means the C no longer equals 0. So we get our Mr. squared plus CM plus K = 0. We solve for our R in order to get our X complement. So we get our XC equaling E to the negative C / 2 Mt times the quantity C1 cosine the square root C 2 - 4 MK over two m * t + C two sine the square root C two C ^2 - 4 MK over two m * t. Now if we find our trial our XP, we'd get a cosine Omega t + b sine Omega T. Taking our first and 2nd derivatives and substituting it back into the original so we can solve for our A and our B. We get 2 equations of negative a * m Omega squared plus BCW plus AK equaling F naughty and negative BM Omega squared minus AC Omega plus BK equaling zero. Once again, our X equation is going to be XC plus XP. So if we look at an example, we get X double prime plus 8X prime plus 25X equaling 200 cosine t + 520 sine T given the initial conditions of X not equaling -30 and X prime of 0 equaling -10. So we're going to start by finding the complementary function, the XC, and we're going to say r ^2 + 8 R Plus 25 = 0, solve for R So we get our XC equaling E to the -4 T times the quantity C1 cosine 3T plus C2 sine 3T. Now if we go back and we look at our X trial, we can see that our X trial would be a cosine t + b sine T, And when I look at the X complement in the X trial, there is no overlap. So then we're going to take our X trial and find the first derivative and the second derivative. Taking those derivatives in the original into the equation that was given and solved for A&B, we can see that A equal 1 and B equal 22. So this new equation is called the steady periodic solution XSP steady periodic solution. So we can see XSP equal cosine t + 22 sine T. Now recall that we have another way of writing this that we can have using our trig identities. If we figure out sqrt 1 ^2 + 22 ^2 and then have that multiplied by cosine of the angle, in this case T minus the tangent inverse of our b / a, they're both positive, the 22 and the one, so we know that we're in our first quadrant. So tangent inverse of 22 / 1 is 1.5254. So our X SP or our steady periodic solution is sqrt 44185 cosine the quantity t -, 1.5254. Now we're actually going to find our X of C So we're going to now go back. We have our X of SP up here, so our X of C ETA the -4 TC one cosine 3T plus C2 sine 3T. We're going to find our X. And our X was the addition of the XC and the study periodic solution, so we're going to add those together. I'm actually going to use the cosine t + 22 sine T because it's easier to find the derivatives of that than the second one. So our X sub T equals this. The two functions added our X prime. We find the derivative using the product rule, etcetera. We put in our initial conditions of X0 negative 30X prime zero -10, and we can find that C1 is -31 and C2 is -52. So our X transient solution TR, our transient solution is going to be E to the -4 T -31 cosine three, t -, 52 sine 3T. Using our trigonometric identities, we're going to take sqrt -31 ^2 + -52 ^2 and that's going to be the square root of 3665 E to the -4 T cosine. The angle, which is 3T in this case minus the tangent inverse of -52 / -31 negative 52 / -31 is going to give us a positive number, but we need to realize it's in the third quadrant because that's when sine and cosine are both negative. So if we're in the third quadrant, we're going to take whatever that number is that comes out and add π to it. So we get 4.1748. So we have found our transient solution. We have found the steady periodic solution and then we found the X for the general solution that works on the given conditions. Thank you and have a wonderful day.