Initial Value Problems ...

Question

Initial Value Problems

Solve the initial value problems in Exercises 89–92.

d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1

Accepted Answer

Welcome back everyone. Solved the initial value problem. The 3rd derivative of R with respect to t is equal to sinc. With the initial conditions R 0 equals 1, R 0 equals 0, and R at 0 equals 2. So for this problem we can begin by finding the second derivative rt, and let's remember that it's basically the integral of the third derivative. So we want to integrate sin cdc and the integral of sin is equal to negative cosinet, right, plus a constant of integration c. Essentially this is an indefinite integral, right? So we are adding a constant for now let's call it C1 because we will have multiple constants and what we can do is simply use the initial condition R 0 equals 1. Which essentially means that on the left hand side we're going to set our derivative equal to 1 and on the right hand side we're going to replace t with 0, so we get 1 equals negative cosine of 0 plus c1. Now cosine of 0 is equal to. 1, so we get 1 plus c1 equals 1. If we add 1 to both sides, we can now show that c1 is equal to 2. So the second derivative r t is equal to negative cosinet + 2. Well done. So we have the 2nd derivative. Now, let's continue and let's identify the first derivative, R T, which is the integral of the 2nd derivative. We want to integrate negative cosinet plus 2 with respect to t. The integral of cosine is sine, right? And we get that negative sign in front, so that the negative sine. C plus and the integral of 2 dt is going to be 2 c and we're going to add a new constant of integration c2. We know that our prime of 0 equals 0, so we're setting the left hand side equal to 0 and on the right hand side t becomes 0, so we get negative sign. Of 0 + 2 multiplied by 0 + c2. Now sin of 0 is 0. 2 multiplied by 0 is 0, so we get 0 equals c2, or basically our c2 equals 0, and that means the first derivative r oft is going to be equal to negative sin t plus. 2 T. Plus 0. There's no necessity to write 0, so that's it. And finally, we can identify R of T. By taking the integral of the first derivative, we want to integrate negative sin t + 2 tdt. Now the integral of sine is negative cosinet, and we have that negatives sign in front, so negative of negative cosinet gives us cosinet plus 2 multiplied by the integral oft, which is t square divided by 2, and we're going to add the constant c3. Notice that we can simplify it slightly. That be cosine c plus t2 plus c3 and we can now identify the constant c3. We're going to set the left hand side equal to 2 because r of 0 equals 2 and on the right hand side t becomes. 0. So we get 2 equals. Cosine of 0. Plus 02 + c3, or basically 2 equals 1 because cosine of 0 equals 1. Less C 3. And finally, C3 is going to be 2 minus 1, which is 1. So we can now show that r oft is equal to cosinet plust 2. Plus one, which is our final answer for this problem. Thank you for watching.

Initial Value Problems Solve the initial value problems in Exercises 89–92. d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1

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Initial Value Problems
Solve the initial value problems in Exercises 89–92.
d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1