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One way to define hyperbolic functions is by means of

Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider ISBN: 9780321747730 43

Solution for problem 46E Chapter 4.2

Fundamentals of Differential Equations | 8th Edition

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Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Fundamentals of Differential Equations | 8th Edition

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Problem 46E

PROBLEM 46E

One way to define hyperbolic functions is by means of differential equations. Consider the equation y’’ – y = 0. The hyperbolic cosine, cosh t, is defined as the solution of this equation subject to the initial values: and The hyperbolic sine, sinh t, is defined as the solution of this equation subject to the initial values: and

(a) Solve these initial value problems to derive explicit formulas for cosh t, and sinh t. Also show that  cosh t.

(b) Prove that a general solution of the equation y’’ – y = 0 is given by y = c1 cosh t + c2 sinh t .

(c) Suppose a, b, and c are given constants for which ar 2 + br + c = 0 has two distinct real roots. If the two roots are expressed in the form  and  , show that a general solution of the equation

(d) Use the result of part (c) to solve the initial value problem: -17/2.

Step-by-Step Solution:

Solution:

Step 1:

In this problem, we have to solve the initial value condition and solve the equation .The hyperbolic functions given y’’ – y = 0  where the hyperbolic cosine, cosh t, is defined as the solution of this equation subject to the initial values and the hyperbolic sine, sinh t, is defined as the solution of this equation subject to the initial values

(a) Solve these initial value problems to derive explicit formulas for cosh t, and sinh t. Also show that  cosh t.

Explanation:

 We have differential equation y’’-y=0

  Now we will assume that the general solution of this equation is

 y(x)=ao+

This the taylor series expression of the y(x)

Now we differentiate this solution with respect to ‘x’

==a1+2a2x+3a3x2………

Again differentiate

==2a2+6a3x………

                                     =2a2+=ao+

                                  =(2a2-ao)+xi=0

                                2a2-ao=0

                                2a2=ao

Therefore we can write ai+2=

Case 1: y(0)=1 and y’(0)=0

        a0=1 and a1=0

       a2==

       a3==

        a4==

       a5==

Hence, we can say that all odd terms are 0 and the even terms are given by

                    a2n=          n=1,2,3…….

Therefore we can write

y(x)=1+++...............

Case 2: y(0)=0 and y’(0)=1

               a0=0 and a1=1

               a2==

               a3==

               a4==0

                 a5==

Hence, we can say that all even terms are 0 and the odd terms are given by

                    a2n+1=          n=1,2,3…….

Therefore we can write

     y(x)=x+++...............

Therefore we can write as

      y(0)=1 and y’(0)=0

                  y(x)=1+++...............

y(0)=0 and y’(0)=1

                  y(x)=x+++...............

Now we know that

       ex=1+x+++....(1)

       e-x=1-x+-+.....(2)

Adding equation (1) and (2)

We get

ex-e-x=(1+x+++...)+(1-x+-+...)

     =2(1+++....)

ex-e-x=(1+x+++...)-(1-x+-+...)

        =2(1+++....)

But (1+++....)=f(x) when f(0)=1 and f’(0)=cos h x

and x+++...............=f(x) when f(0)=0,f’(0)=sin hx

Hence , 2 cosh x=ex+e-x

             cos hx =

Similarly

         2 sinh x=ex-e-x

             sinhx =

Now , (cos hx)=()==sin hx

Similarly

(sinhx)=()==cos hx

(b) Prove that a general solution of the equation y’’ – y = 0 is given by y = c1 cosh t + c2 sinh t.

 Explanation:  y’’ – y = 0

 Characteristic equation will be x2+0x-1=0

     or  x2-1=0

 Roots =1,-1

Therefore, the general solution to this given differential equation is Aex+Be-x

We can write (+)ex+(-)e-x

                      (ex+e-x)+  (ex-e-x)

                      (A+B)cos hx + (A-B)sin hx

      General solution: C1coshx+C2sin hx

(c) Suppose a, b, and c are given constants for which ar 2 + br + c = 0 has two distinct real roots. If the two roots are expressed in the form  and  , show that a general solution of the equation

Explanation:

ar2+br+c has two real roots (

b2-4ac>0

(

(

The general solution of the given equation

                       = A1e(+   A2e(

                     =e( A1e(+   A2e()

Now , e(=(cos h +sin h) and  e(=(cos h -sin h)

  General solution:

y=e( A1e(+   A2e()

=e(A1(cos h +sin h

Step 2 of 1

Chapter 4.2, Problem 46E is Solved
Textbook: Fundamentals of Differential Equations
Edition: 8
Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider
ISBN: 9780321747730

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