Determine (without solving the problem) an interval in which...

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.\\ $(16 - t^2)y' + 3ty = 7t^2$, $y(-7) = 1$ \\ ___ $< t <$ ___

Transcript

00:01 Hello and welcome to problem 15 of chapter 2 section 4.

00:03 Here we're asked to solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value of why not.

00:11 So our given initial value problem here is y prime plus y cube is equal to zero.

00:17 We can simply move the y cubed over and divide it.

00:22 So we'll have d .y d t is equal to negative y cubed.

00:30 From here, we can bring the y down and the d t up and simply integrate.

00:34 We'll be left with the d y over negative y cubed is equal to d t.

00:43 So i'm going to bring this negative to the other side and drop that.

00:48 From here we can just integrate, integrate the sides.

00:55 Great, so we'll have the integral of d y times y to the negative 3.

01:01 What is the integral of y to the negative 3 d .y.

01:09 Well, that's y to the negative 2 over negative 2.

01:16 And that's going to be equal to the integral of negative t, which is negative t plus c.

01:27 Okay.

01:28 And from here, we can just solve for y, and we'll do that by multiplying everything by negative 2.

01:36 So actually, let's move this up here.

01:39 We'll have y to the negative 2 is equal to, well, 2c plus c.

01:49 And remember, this is actually 1 over y squared, 1 over y squared.

02:00 So we're going to bring this 2t plus c term down, the y squared term up, and then take the square root.

02:06 So lastly, we'll be left with y is equal to plus or minus the square root of 1 over 3.

02:14 2 t plus c and that is our initial those the solution to our initial value problem however we do have this initial value so we'll plug that in let's erase down here a little bit so if we plug in our initial value be left with to green y y -not is equal to plus or minus the square root of one over c where we can take the square we can square both sides we'll get y -not squared is equal to one over c such that c is equal to one over y -not squared great and from here we can yeah from here we can substitute this one over why not squared term for c back in to our solution and it's right here we'll have c and then we'll have one over why not squared great and looking at this there are we have to find out where uh sorry what why not cannot be and um just by simply looking at it we know that we cannot divide by zero so why not cannot be equal zero let's uh let's write that done here y0 cannot be equal to zero.

04:07 Let's also just rewrite the solution.

04:10 So we'll have y is equal to plus or minus square root 1 over 2t plus 1 over y not squared...

Question

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.\\ $(16 - t^2)y' + 3ty = 7t^2$, $y(-7) = 1$ \\ _ $< t <$ _

Question

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.\\ $(16 - t^2)y' + 3ty = 7t^2$, $y(-7) = 1$ \\ ___ $< t <$ ___

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Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.\\ $(16 - t^2)y' + 3ty = 7t^2$, $y(-7) = 1$ \\ _ $< t <$ _