4.1.5

[STARTS AT: 2:53]

Speaker: Now, for example five, we don't have a proposed solution, but this one is set up nicely for us. We've already got y prime equals, so we don't have to isolate anything. So let's use the antiderivative. We're going to find the antiderivative. 

Now, I would much prefer-- I'm going to rewrite this as y-- as dy equals 3 e to the x plus x squared minus 4 dx. That's how I want to write this. And I'm going to take the antiderivative with respect to x on the right, with respect to y on the left. This would be y equals 3e to the x plus x cubed over 3 minus 4x plus c. 

Well, that's great, but now I need to use my initial value. I'm going to say y of 0 equals 5. So I have 5. Since we have that, you see that 5 equals 3 times e to the 0 plus 0 cubed over 3 minus 4 time zero plus c. That would be 3, 0, 0 so 3 plus c equals 5, and c equals 2, which makes our particular solution y equals 3-- yeah, 3 to the x. Plus, I'm pulling that from-- I'm pairing these two statements together, 3 to the x plus x cubed over 3 minus 4x plus 2 as our particular solution.