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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.$$y(x)=c_{1} \cosh 3 x+c_{2} \sinh 3 x, \quad y^{\prime \prime}-9 y=0$$.

Solution is defined on interval $(-\infty, \infty) .$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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in this problem, we are given a differential equation over here and a possible solution. Why of X and we want to make sure that this actually is a solution to this differential equation. First, to do that, we need to make sure that why Vex is at least twice differ in trouble. And we can see that it is because it's a function of hyperbolic sign and hyperbolic co sign which we're gonna be infinitely differential so we can go ahead and take these derivatives here. So why Prime of X is equal to the derivative of hyperbolic six. Co sign is just hyperbolic sign. But we have to use the chain role and bring this three out to the front. So we get three c one hyperbolic sine of three x plus similar thing over here. Three c two hyperbolic co sign the three X and then we can take the second derivative my double prime of X equal Teoh nine c one hyperbolic co sign the three X plus nine C two hyperbolic sign of three FX. Then we want Teoh. Take these equations that we found. We found this quiet double prime and we have this y of X and plugged them into our differential equation over here and make sure we get zero. So let's do that. Why? Double prime is nine C one hyperbolic co sign of three X plus nine. See to hyperbolic sine three X and then we want to subtract off. Nine. Why so minus nine? See one code hyperbolic co sign of three X plus C two hyperbolic sign of three X. That's three X. Okay, then we see we can go ahead and factor than nine out of this and we get nine. See one hyperbolic co sign three x plus, See too hyperbolic sine three x minus nine of the same thing. Um so just declared by this was just simplifying this part. And so we see we have nine c one hyperbolic coastline of three x plus c too hyperbolic sign of three X minus nine of the same thing. This is equal to zero. And that means that yes, this way of X is a solution to this differential equation. The last thing we need to check is for what X? Is this a solution? Well, hyperbolic sign and co sign, um, will take any X you can plug any real Number X into those functions, and all of our derivatives are just some, uh, some functions of hyperbolic sign and co sign. So we get any X in the real numbers will work or any X between negative infinity and infinity.

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