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Solve the following initial-value problems by using integrating factors.$$x^{2} y^{\prime}=x y-\ln x, y(1)=1$$

$y=\frac{\ln (x)}{2 x}+\frac{1}{4 x}+\frac{3}{4} x$

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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this video. We are going to solve this differential equation. Why? Prom minus y o X equals minus lock eggs over x square with the initial condition that when X is one, Why is one s will? So we have a rented into the standout form already. We know that p of X is minus one over eggs so we can find integrating factors as he to the minor locks. Absolute X wish is one over absolute X. Now, when we multiply this true on both sigh, we can consider the absolute value as just pulling a meal because we have absolute value on both side of the equation. In the case that give miners were Lou off them gonna give my nuts? Why, Lou? So we can cancel the minus one. How? From both sides So we can treated like this. Asked just no mobile e, no meal. Now, when we modified to the left side gonna collapse into this form as usual. What's left is to integrate this thing on the right hand side. If we can integrate this, we gonna have that. Why use eggs times that thing that'd be integrated. So let's do it here. I pull minus one out. We were inserted back later on. What we have to do is is obviously this style this home some take the eagle, buy pots used to sew everything. Um, so you will be long off eggs and then we will be one over X Q The ex. So do you will be. Why no eggs the eggs and we will be minus one over two x square. And so we have all the ingredients. We just put them in here. You we will be minus lock eggs over two x square And here we might ask to minus become plus one over two X cube now D X, And this is just pulling a meal so we can integrate it Fly away. It will become minus one over four X square. And don't forget the Constanta. Yeah, I would like to write as minors so that when we multiply minus one, true this everything will become Plus, it doesn't matter for for the constant is I can be whatever. Now what we have to do, we lock that in and we gonna get Yeah, I skip some steps. That is B I Q X through everything in here. We're gonna get long X over two X plus one over four X says, see times, eggs. But it is not over. We have initial condition, Right? So x equal one, Gibbs. Why equal one as well? We have one equal lock off. One year, zero and this one over four. And this is C. So our constant room here will be tree or were four, right? And so we have. So this problem, we just inside this bag, this right here. So the answer to this particular equation is lock X over two eggs, plus one over four eggs, plus tree over four times eggs. And that is the answer. Thank you.

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