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Solve the given differential equations.$$x d y+y d x+x d x=0$$

$y=-\frac{x}{2}+\frac{c}{x}$

Calculus 2 / BC

Chapter 31

Differential Equations

Section 3

Integrating Combinations

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Section 16 that to problem number ones that were dealing with first order, ordinary, different journal equations in this particular section. So what have instructed us that it was to try to get everything we can into the form of delight. Our standard form is D Y DX, plus some expression or function of X. Times Y is equal to Q rex, so if we can get into this form, then we can multiple. We can manipulate the left side to be the derivative of a product by multiplying by an integrating factor. So they tell us that the integrating factor that we multiply both sides of the equation so the integrating factor will be e to the integral of p of x dx. So let's get this in standard form. If I divide everything by X, I get D Y d X plus one over X times. Why is equal to eat to the X over X? So my function p of X is just one over acts. So the integrating factor here, it's just going to be e to the integral one over x dx. And since I know that excess positive, this is just e to the natural log of X, which is just X So that is my integrating factor. So take the equation that we have de y the X plus one over X. Why equals et the X over X? Multiply it by that integrating factor. The integrating factor being X So I multiplied by X. Then what? I end up with this X times Do I the X plus why is equal to e to the X? I'm back to my original. But what you should recognize is that this expression on the left that is the derivative with respect to X of x y. Because if you were to differentiate X, why you would use the product rule you have x times, the derivative of y plus the derivative of X times. Why so x t y dx plus why? Um so that's what you see here. So the derivative of X Y is equal to he to the X. So let's just rewrite this. So I've got the derivative with respect. Tax of X Y is equal to e to the X. Now I can integrate both sides of this equation and I get X Y is equal to the integral of eat of the X, the X That means that X Y is equal to eat of the X plus a constant of integration. Therefore, why is equal to e to the X over X, um, plus C over X. Now you could leave in like that. You could write, you know x to the minus one into the X plus c x to the minus one. Either way, so either these answers will be fine for the answer to this differential equation.

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