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.

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