One way to define hyperbolic functions is by means of

QUESTION:

One way to define hyperbolic functions is by means of differential equations. Consider the equation \(y^{\prime \prime}-y=0\). The hyperbolic cosine, cosh t, is defined as the solution of this equation subject to the initial values: \(y(0)=1\) and \(y^{\prime}(0)=0\). The hyperbolic sine, sinh, is defined as the solution of this equation subject to the initial values: \(y(0)=0\) and \(y^{\prime}(0)=1\).

(a) Solve these initial value problems to derive explicit formulas for \(\cosh t\), and \(\sinh t\). Also show that \(\frac{d}{d t} \cosh t=\sinh t\) and \(\frac{d}{d t} \sinh t=\cosh t\).        

(b) Prove that a general solution of the equation \(y^{\prime \prime}-y=0\) is given by \(y=c_{1} \cosh t+c_{2} \sinh t\).

(c) Suppose a, b and c are given constants for which \(a r^{2}+b r+c=0\) has two distinct real roots. If the two roots are expressed in the form \(\alpha-\beta\) and \(\alpha+\beta\), show that a general solution of the equation \(a y^{\prime \prime}+b y^{\prime}+c y=0\) is \(y=c_{1} e^{a t} \cosh (\beta t)+c_{2} e^{\alpha t} \sinh (\beta t)\) 

(d) Use the result of part (c) to solve the initial value problem: \(y^{\prime \prime}+y^{\prime}-6 y=0, y(0)=2, y^{\prime}(0)=-17 / 2\)

Equation Transcription:

Text Transcription:

y prime prime - y = 0

y(0) = 1

y prime (0) = 0

y prime (0) = 1

y(0) = 0

cosh t

sinh t

d/dt cosh t = sinh t

d/dt sinh t = cosh t

y prime prime - y = 0

y = c_1 cosh t + c_2 sinh t

ar^2 + br + c = 0

alpha - beta

alpha + beta

ay prime prime + by prime + cy = 0

y = c_1 e^alpha t cosh(beta t) + c_2 e^alpha t sinh(beta t)

y prime prime + y prime - 6y = 0, y(0) = 2, y(0) = -17/2