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Verify that the given function is a solution to the given differential equation $\left(c_{1} \text { and } c_{2}\right.$ are arbitrary constants), and state the maximum interval over which the solution is valid.$y(x)=c_{1} e^{a x}+c_{2} e^{b x}, \quad y^{\prime \prime}-(a+b) y^{\prime}+a b y=0$where $a$ and $b$ are constants and $a \neq b$

is the solution

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

Campbell University

Oregon State University

Harvey Mudd College

Boston College

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in this problem. We are given a differential equation over here, and we want to check that this way of X is a solution to this differential equation. So to do so, First, we need to check that. Why of X is at least twice differential, which it is. It's a function of addition and exponential functions, so we can go ahead and take the derivative of this wide back. So my prime of X is equal to a times C one e to the A X plus be in time. See, too times e to the B X, and we can take the second derivative. So why Double Prime of X is a squared times C one e to the a X plus B squared times See two times e to the B X. All right, so now we have expressions for why, Why prime and why Double prime and we can plug them into our differential equation and make sure that we get zero. So it's going to do that. Why Double prime is a squared time. See one e to the a X plus B squared times 32 times. Eat the B X, then we want to subtract a plus B times by pine, which is a time see one times E to the A X plus B times C two times E to the B X, and then we want to add a Times B times. Why so, plus a B times are expression for Why is C one e to the A X plus C two times e to the B X. All right, so we can go ahead and simplify this out and get a squared times. See one times e to the A X plus B squared time. See two times e to the B X minus. We will go ahead and distribute this a plus B so a squared see one times e to the A X minus. B times a times C one times E to the A X um, now minus a times B from sea to times E to the B X and minus B squared C two times. Eat to the beat X. Now we just need to distribute this last a times B to this term. So plus a Times B Times C one terms each. The a X plus a Times B Times C two times e to the B X and we can see some cancellations here. So we have an A B C two tons of the BX and here we have a negative ABC to eat. The BX slows cancel. We also have an eight time speed terms, see one times either the a X, which cancels with negative B times a temp C one tons to the a X Um, we have a positive B squared temp C two times of the B X and a negative B squared times. See two times with the X and finally in a square Dempsey, one times eat of the a X and a negative a square Tennessee, one times eat of the X So all of our terms cancel and we end up with zero. So, yes, this way of X is a solution to this differential equation. And the last thing we need to do is check for which X this is could be true. Well, are only, um I think we need to check is that we can plug in any x two y of x y part of X and y double prime Vex. And there's no reason that we couldn't plug in any real numbers, since we can plug in any real X two exponential functions. So we get that X is in any real number or that X is in between negative infinity and infinity.

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