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What are the integrating factors for the following differential equations?$$y^{\prime}=3 x+x y$$
$y=-3+C e^{\frac{x^{2}}{2}}$
Calculus 2 / BC
Chapter 4
Introduction to Differential Equations
Section 5
First-order Linear Equations
Differential Equations
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section 4.5 problem 227. So we're dealing with ordinary differential equations that can be written in standard forms. If I can write an equation like this, why prime plus some function of X y equals some function of X? Then I can integrate this using an integrating factor. So let's get this in standard form. So it's going to be why prime minus X y equal three x So the integrating factor is going to be e toothy integral minus X dx. So this is going to be e to the minus. X squared over two is the integrating factor. So I need to multiply both sides of this equation by the integrating factor. So when I do that, why prime minus X y equals three acts? I won't buy all of that by e to the minus X squared over two. What you end up on the left side of the equation is just y e to the minus X squared over two. Prime is equal to three x e to the minus X squared over two, and then we can go ahead and look and say, Okay, what would it take them to integrate both sides of this equation. So when you integrate both sides of this equation on the left side of the equation, you're going to get why e to the minus x squared over two? Uh huh. And then what we have on the right side of the equation? It might help if I write this as, um three. Then I could write it as, what, minus two x. So put a negative in front of their this and he to the minus. Excuse me? X squared over two. So if I did appreciate that um, good. Um, just minus X, sir. Let me back up. Sorry. Let's do it like this. So I'm trying to do too many steps at one time. So what I can do here is make a substitution. If I let you equals e to the minus X squared over two, then do you is e to the minus. So that's minus two X over to e to the minus. X. Um, pardon me. I'm every day, different day here. So if you differentiate that you're going to get e to the minus X squared over two when you different shape minus x squared over to you get minus X. So what you see here is minus X dx and that in a row. So what this becomes is why e to the minus X squared over two is equal to minus three. And this is just the integral of e to the u. You did that you do you? So this is just minus three. He today you plus a constant of integration. So my answer here is why e to the minus x squared over two. The quota minus three you to the minus X squared over two plus a constant of integration. Now you divide everything by e to the minus x squared over to you get wise, you know, minus three plus c e to the X squared over two that is the family of functions that serve as the solution for this different equation.
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