Hyperbolic functions are useful in solving differential...

Hyperbolic functions are useful in solving differential equations (Chapter 8 ). Show that the functions $y=A \sinh k x$ and $y=B \cosh k x,$ where $A, B,$ and $k$ are constants, satisfy the equation $y^{\prime \prime}(x)-k^{2} y(x)=0$

Key Concepts

Linear Homogeneous Differential Equations

Linear homogeneous differential equations with constant coefficients are equations where each term is a constant multiplied by the function or its derivatives, and the sum equals zero. Such equations often have solutions that are exponential functions or combinations of hyperbolic and trigonometric functions. Verifying a proposed solution involves substituting it and its derivatives into the equation to ensure the equality holds.

Differentiation of Hyperbolic Functions

The differentiation properties of hyperbolic functions are essential in verifying solutions to differential equations. Specifically, the derivative of sinh(x) is cosh(x), and the derivative of cosh(x) is sinh(x). These relationships allow for straightforward computation of higher-order derivatives necessary for substituting back into differential equations.

Hyperbolic Functions

Hyperbolic functions, including sinh and cosh, are defined using exponential functions and serve as analogs to trigonometric functions. They are frequently used in solving differential equations because their exponential nature makes them natural solutions to equations with constant coefficients.

Transcript

00:01 All right, so we've got hyperbolic functions which we're claiming, and i agree, are useful in solving differential equations.

00:09 So let's go ahead and do that real quick.

00:11 First of all, what are hyperbolic functions? well, sine h of x is equal to e to the x minus e to the minus x over 2, and cosh of x is e to the x plus e to the minus x over 2.

00:27 What are the derivatives here? well, sine h prime of x is equal to, let's see, by linearity we've got just e to the x up here, minus e to the minus x, so this minus comes up, makes us a plus, plus e to the minus x over 2, and cosh of x is, again, e to the x's own derivative, this minus falls forward, becomes plus into a minus, e to the minus x over 2, which we'll notice are respectively cosh x and sine x.

01:01 So they have this kind of reciprocal relationship, sine h differents to cosh, which differentiates to sine h.

01:12 That's very fun and nice.

01:16 So with that in mind, let's show that why a sine h kx and b cosh kx both satisfy this.

01:25 Well, let's see.

01:27 We'll first start with sine h double prime of kx, i guess we'll a times all that, minus k squared times sine h of kx, times a of course.

01:47 I'd like to show what this is equal to.

01:48 Well, sine h double prime is going to become cosh and then sine h again, but by the product rule you have to take a factor of k out each time.

01:56 This becomes a times k squared times just sine h of kx minus, we'll flip this around, a times k squared sine h of kx...

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