4.5.2
[STARTS AT 5:02]
Professor: All right, now, for number three, we're going have an initial-value problem, as I was just alluding to, y prime plus 3y equals 2x minus 1 where y of 0 equals 3. Now, the first thing we have to do for this one is make sure it's in standard form. It is.
So we'll go ahead and identify what p of x and q of x are. But we need p of x to find our integration factor. So p of x is 3 and q of x is 2x minus 1.
So now coming over here, I want to find mu of x, that's going to be e to the antiderivative of 3dx. So that will be e to the 3x.
Now, notice we sort of ignore the constant. Here, really there should be a constant. But because we end up with a constant at the end of our questions, it really doesn't change anything, so it may not be completely correct. But it's appropriate to leave off the plus c's on the integration factor.
So now that we have this, step three is multiply-- try that again-- multiply our equation, both sides of that equation, by our integration factor. So we're going to multiply e to the 3x on both sides of this. So this will be e to the 3x y prime.
Actually, we may want to-- yeah, let's go ahead and do that-- plus 3 ye to the 3x equals e to the 3x. Now, this side I'm going to leave factored. I want to expand it on the left side because that shows me that this is the derivative of e to the 3x times y.
OK, all right, so we are now up to step three. Now, step four would be to take the antiderivative of both sides. Again, with respect to x, left hand side fundamental theorem of calculus tells me that is going to be e to the 3x times y. And the right side, well that's actually going to need integration by parts, integration by parts.
So we would say, OK, u and dv, and our u, a good choice for that would be 2x minus 1, because the derivative of that is 2. And e to the 3x, the integral of that is 1/3e to 3x. So this is going to be equal to u times v. So 1/3e to the 3x is the antiderivative of vdu.
So we get minus 2/3 e to the 3xdx, which would be minus 2/9e to the 3x plus c. And we'll go ahead and write this. It's a fraction.
All right, last step, step five. Divide by e to the 3x, which gives us y equals 2x minus 1 over 3 minus 2/9 plus c over e to the 3x. Actually, we have one more step. We need to determine what the constant is, a constant of integration.
We do have an initial value though, of y of 0 equals 3. And we know the e to the 3 times 0 is going to be 1, so this should be 0-- so that would be negative 1/3 minus 2/9 plus c equals 3, which means that c is equal to 32/9, which means our particular solution is 2x minus 1 over 3 minus 2/9 plus 32/9e to the 3x.