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# Use the method of Exercise 23 to solve the differential equation.$xy' + y = -xy^2$

## $$y=\frac{1}{x \ln |C x|}$$

#### Topics

Differential Equations

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##### Catherine R.

Missouri State University

##### Samuel H.

University of Nottingham

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### Video Transcript

in order to solve this. The first step is to divide both sides by acts and all the terms by axe nor jet into standard form with pee vacs and cue of X. So doing this, we know we can end up with D y over D ax plus one over X times. Why is negative one? Why squared? You can clearly see now we have d Y over DX plus p of x times. Why is Q backs times wide? Toothy? End other words. This was a good step because it's not put this into standard form proving twist that r p of X is one over acts, which means now we know you is why the one minus on. Which means now this could be simplified to why to the one minus two, which is wide to the negative one or one over Why? Which means we now multiplying this by each of the terms D'You over D axe is negative. One over y squared d y over d backs substituting. Do you are over DX with negative y squared times Do you over DX do you over D axe? Okay, remember Are you in this context? Was won over. Why? So we're simply substituting Okay? Time. Look at the interpreting factor. We have eats the integral of negative one over axe dx again. This came from our p of X earlier. Integrating this we know Eat the natural log is simply going to give us one, which means this could be crossed out, which means we simply end up with one over. Axes are intruding factor. Multiply each of the terms by this. Multiply each of the terms by the intruding doctor. Just what I'm doing right now you can use the product roll differentiation the integral of one of her acts. DX. We're now going to be doing that as our final stop. We get the natural log of the absolute value of X plus not for work of C. Lastly, remember, you actually has a value of one over. Why? So club got back and and then, lastly, why must be in terms of itself, we can't have a coefficient of one over X in front. It must simply be why equals something? That's what they imply when they mean solution. So right, this simply in terms of why

#### Topics

Differential Equations

Lectures

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