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Solve the following initial-value problems by using integrating factors.$$\left(1+x^{2}\right) y^{\prime}=y-1, y(0)=0$$

$y=1-e^{\tan ^{-1}(x)}$

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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section Ford up five Problem. 245 Here we're dealing with the near first order ordinary differential equations and I need to get this equation in standard form. So why prime plus some function of X times Why equal some other function of X? Then I will multiply by integrating factor which is just e to the anti derivative p of x. So to get this in standard form, this is weird to be why prime minus one over X squared plus one. Why is equal to negative one over X squared plus one? So my integrating factor is going to be e and then minus one over x squared plus one x the anti derivative there That's just the arc tangent. So this is e to the minus our tension of X. So that is my integrating factor. So this equation reduces down to why times e to the minus arc tangent of X prime is equal to minus one x squared plus one and e to the minus arc tangent of X. Now it just becomes a matter of integrating both sides of this equation trivial. On the left side, it was the whole reason for a designing on this. So this is just going to be why and then e to the minus arc tangent of X. And then when I look at the right side of the equation can make a substitution here, let you, um, equal minus the art tangent of X Then d you going to remind us one over expert plus one DX. So in reality, what you see here with this particular integral, this is just simply the integral of e to the u Do you Ah, in that particular form. So this is going to be when you integrate that you'll get e to the U plus a constant of integration. And so that's just e to the minus our tension of X plus a constant of integration. So now, in order to solve for why you divide by e to the minus are tended of X, so you will get why is equal to So when I look at this So when I don't fight this out, um, uh, you get y is equal to one plus see e to the arc tangent of X. And then I had an initial boundary condition. Wives zero is zero So why zero zero? That means that zero is equal to one plus. See e. The Arc Tangent of zero is just simply zero. So this means that C is equal to negative one. So my final answer is why is equal to one minus e to the arc Tangent of X. That's the final answer here.

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