00:01
So here we want to solve this initial value problem.
00:06
So we're given the differential equation, y prime, is equal to y cubed sine of x.
00:21
Okay.
00:23
And we're given the initial condition, we're given that y of zero is equal to zero.
00:31
Okay.
00:32
So, well, we have y prime.
00:36
So y prime is the derivative of y, so that's d y, d x, okay? and that is equal to, well, y cubed, y cubed sine of x.
00:52
This is equal to y cube sine of x, just writing y prime as the y, d x.
01:00
Okay so then we have well separating our variables we have that d y over y cubed right just implies that d y over y cubed is equal to well sign of x d x x is multiplying both sides by basically d x and then divide them through by y cube so we get d y over y cube is equal to well sign of x d x d x okay then we can integrate both sides.
01:35
So we have the integral of, well, d .y over y cubes.
01:41
That's the integral of 1 over y cube.
01:44
So the integral of 1 over y cubed, d .y is equal to the integral of sine of x dx.
01:57
Okay.
02:03
So what we get, well, the integral of 1 over y cubed, get the integral of y to the, oh, this should be, yeah, this is, yeah, that's right.
02:15
This is the integral of one over, well, the integral of one over y cubed is the integral of y to the negative three power, right? so what we get here, when we integrate, we get y to the negative, well, to the negative three plus one.
02:41
Which is a negative, well, so it's negative 2, but negative 3 plus 1, over, well, over negative 3 plus 1, right? and that is equal to the integral of sine of x, well, that's negative cosine of x, plus a constant, okay? so then using, let's see, so yes, i mean, this, i mean, again, you put this as y equals, y to the native 2, right? but this is just using the formula, the integral of x to the n, the x is equal to what x to the n plus 1 and then we divide through by n plus 1, right? okay...