Solve the initial value problems in Exercises 71–90.
y⁽⁴...

Question

Solve the initial value problems in Exercises 71–90.

y⁽⁴⁾ = −sin t + cos t;

y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

Accepted Answer

Welcome back everyone. In this problem, we want to solve the initial value problem that says the 4th derivative of y equals 2 sin t plus 3 equals t, with the 3rd derivative of y at t equals 0 equals 0. The 2nd derivative of y at T equals 0 equals 1. The first derivative of y at T equals 0 equals 2, and the value of y at T equals 0 equals -2. Now in this problem, basically, if we're solving the initial value problem, what we're trying to do is to find our initial equation, our initial value for y, OK, and to do that, given the 4th derivative, that means we'll basically need to integrate 4 times to get our value for y, and then we can figure out our constants using our values of y given t equals 0. So let's go ahead and do that. So for starters, OK, we'll need to integrate, OK, to get our 3rd derivative, we'll need to integrate our 4th derivative, OK. Integrate our 4th derivative with respect to t. So that is going to be the integral of 2 sin, let me write that properly here, 2 sin t plus 3 cost t. With respect to tea. Now we know in integration it means that we can integrate both of these terms since we're finding their sum. So really this is the integral of 2 sint or 2 multiplied by the integral of sin t with respect to t, plus 3 multiplied by the integral of cost t with respect to t. And now when we go ahead and solve here. Then we get it to be equal to -2 + t, the integral of 2 multiplied, or sorry, which is the result of 2 multiplied by the integral of sin t with respect to t. Plus 3 sint because the integral of cost t is sint plus C1, OK? So this is our third derivative of Y. Now for us to move forward, we'll need to figure out what constant C1 is here and to do that we can apply, so we can apply the fact that we know our third derivative at T equals 0 equals 0, OK, so we can use this to solve for C1. Because no This would mean then that. I, sorry. Our third derivative at t equals 0, which equals -2 + 0, + 30 plus c1, would be equal to 0. And thus this means then. That if we calculate these cost 0 is 1, so this is -2, 3 sins 0 is 3 multiplied by 0 or 0, so -2 + 0 plus C1 equals 0, which means then that C1 will be equal to 2. So, let me put this in red here. Actually, this tells us then that our third derivative is going to be -2 T + 3 sint + 2, OK. So this is our 3rd derivative. Now we need to integrate this to get our 2nd derivative, so let's do that. Now, our 2nd derivative, like we said, it's going to be the integral of our 3rd derivative, so really, OK. Let me write that here in the proper notation. So really, this is going to be equal to the integral of -2 Castile. Plus 3 sign T, OK, plus 2, all with respect to T. And now when we calculate here. We should get this to be equal to -2 sin t minus 3 cast t plus 2t from the integral of 2 with respect to t is 2 t plus c2. So now again we need to figure out our value of C2 and to do that we can apply the fact that our second derivative when T equals 0 is equal to 1. So we can apply this to solve for C2. So now, when we do that, OK. Then we get our second derivative, which we know is 1. To be equal to -20 minus 3 + 0 plus. 2 multiplied by 0 plus C2. So in that case, then 1 equals, well this -2 sin 0 would be 0, plus 0 is 1, so we know this is -3 + 0 plus C2, so 1 equals -3 plus C2, which means then that C2 must be equal to 4. So what does this tell us? Our second derivative is equal to -2 sint minus 3 T + 2T + 4. And again, I'll just put this in a box to the right just for us to keep track of our values of our derivatives as we go along. Now let's use this to find our first derivative. Now, here, this means then, like we said before, our first derivative is the integral of our second derivative, so it's really the integral of -2 sint minus 3 T plus 2T + 4, OK. And now when we integrate here, we end up with 2 T minus 3 sint plus T 2 + 4T plus some constant C3. Now we can apply, applying the fact that our first derivative at T equals 0 is 2. To solve for our 3rd constant C 3. Because now, in this case. Then that means 2 is going to be equal to 20 minus 30 + 02 + 4 multiplied by 0 plus c3. And now if we simplify here, we'll get 2 to be equal to 2 plus c3, which means then that C3 is going to be 0. And now if C3 is zero, what does that tell us? It means then that our first derivative is equal to 2 t minus 3 sin t plus t2 + 4t. Now we're almost there, we have our first derivative, and remember the integral of the first derivative is the function. So in other words, why, OK, why, let me put it in red actually. Why Is going to be the integral of our first derivative, so it's the integral of 2 t minus 3 t plus T 2 + 4T with respect to T. And now, as usual, if we go ahead and integrate each of our terms, that's going to be 2 t plus 3 cost T. Plus 1/3 of T cubed plus 2 T 2 plus C4, and now all we need to do is to solve for C4. Applying the fact that Y of 0, equals -2, the solve for C4. Then if we put it into our problem here, that means y 0-2 is going to be equal to 2 sin 0 + 30 + 1/3 of 0 cubed plus 2 multiplied by +02 plus c4. I know when we simplify, we should get -2 to be equal to. 3 plus C4, OK, which means then that C4 will be equal to -2 minus 3 or -5. Therefore, The solution to our initial value problem is Y of t. is equal to 2 sin t plus 3 t plus 1/3 of t cubed plus 2 t 2 minus 5. Thanks a lot for watching everyone. I hope this video helped.

Solve the initial value problems in Exercises 71–90.y⁽⁴⁾ = −sin t + cos t;y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

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Solve the initial value problems in Exercises 71–90.
y⁽⁴⁾ = −sin t + cos t;
y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0