Express the solution of the given initial value problem as a...

Express the solution of the given initial value problem as a sum of two oscillations. Primes denote derivatives with respect to time t. Graph the solution function x(t) in such a way that you can identify and label its period. x'' + 49x = 39 cos 6t; x(0) = x'(0) = 0 Express the solution of the given initial value problem as a sum of two oscillations. x(t) = \ (Simplify your answer. Type an exact answer, using radicals as needed.) Graph the solution function x(t). Choose the correct graph below. Identify the period. P = \ (Type an exact answer in terms of $pi$.)

Transcript

00:01 In this question, we want to find the, in this period, we want to solve this differential equation.

00:09 So, in order to do this, we would need to find first the homogeneous solution and add it to the particular solution.

00:23 And for the homogeneous solution, all we need to do is we need to add this to, all we need to do is to set this differential equation equal to zero.

00:34 So that means you have x double dot plus 49x would be equal to zero.

00:40 Given the characteristic roots, s squared plus 49 equals 0 implies that s would be equal to j7 plus or minus.

00:52 So we can very quickly come up with the homogeneous solution because since our roots of the characteristic polynomial are complex, we would have that this is equal to c1 times cosine of 70 plus c2 times the cosine of 17 times the sign of 17.

01:11 Now for the particular solution, we're going to assume that it is going to be of the form a, cosine of 60, plus b, sine of 60.

01:34 We need to do this because their derivatives are linearly independent.

01:42 So we need to take some derivatives now.

01:44 So you have negative 6a sine plus 6b cosine of 60, and then the second derivative of our particular solution would be equal to negative 6a ,000.

01:57 Negative 36 cosine of 60 minus 36a and then this will have a b, 36b sign of 60.

02:12 Now what we need to do is we need to substitute all this information into our differential equation, namely this and this.

02:20 So what we're going to have, when we plug in, we're going to have negative 36a, and i'm going to drop the arguments of the cosine and signs just so that this is easier to write.

02:32 Minus 36b, sine, plus 49a, cosine, plus 49b sign.

02:42 So this will be equal to 36 cosine.

02:53 Notice this is way easier to calculate than we thought it would be.

02:57 If we equate the coefficients, we would have to have 49 minus 36 is 13, cosine of 60 will be equal to 13a, 13a cosine cosine cosine cosine, 60, sorry, 13a, cosine of 60 would be equal to 39, cosine of 60, and 13b, sign of 60 will have to equal 0...

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