Y= c_1e^x + c_2e^-x is a two-parameter family of solutions of the second-order de y'' - y = 0. find a solution of the second-order ivp consisting of this differential equation and the given initial conditions. 11. y(0) = 1, y'(0) = 2 12. y(1) = 0, y'(1) = e 13. y(-1) = 5, y'(-1) = -5 14. y(0) = 0, y'(0) = 0

11)y =

12)y =

13)y =

14)y = 0

Step-by-step explanation:

Given data:

y''-y=0

The equation is

-1 = 0

(m-1)(m+1) = 0

if  above equation is zero then either

m - 1 = 0 or  m + 1 = 0

m = 1        ,    m  = - 1

11)

y(0) = 1 , y'(0) = 2

+   = 1   (y(0) = 1) (1)

-   = 2   (y'(0) = 2)  (2)

adding 1 & 2

2 = 3

= 3/2

3/2 +   = 1

 = 1 -  3/2

= - 1/2

y =

12)

y(0) = 1 , y'(0) = e

+   = 0 (y(0) = 1) (3)

= -

  (y'(0) = 2)  (4)

replace = by equation 3

taking common

∴ y =

13)

y(-1) = 5 , y'(-1) = -5

+   = 5   (y(-1) = 5 ) (5)

-   = -5    (y'(-1) = -5)  (6)

Adding 5&6

2 = 0

= 0

= 5 -

= 5 - 0

= 5/e

y =

y =

14)

y(0) = 0 , y'(0) = 0

+   =  0 (y(0) = 0) (7)

-   = 0   (y'(0) = 0)  (8)

Adding 7 & 8

2 = 0

=

y = 0

stop cheating

step-by-step explanation: