Question

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

Answer 1

11)y =
`(3)/(2) e^(x) - (1)/(2) e^(-x)`

12)y =
`(e^(2) )/(1+e^(2) ) (e^(x) - e^(-x) )`

13)y =
`5e^(-(x+1))`

14)y = 0

Explanation:

Given data:

`y=c_(1) e^(x) +c_(2) e^(-x)`

y''-y=0

The equation is

`m^(r)`-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

`y'=c_(1) e^(x) -c_(2) e^(-x)`

`c_(1)` +
`c_(2)` = 1 (y(0) = 1) (1)

`c_(1)` -
`c_(2)` = 2 (y'(0) = 2) (2)

adding 1 & 2

2
`c_(1)` = 3

`c_(1)` = 3/2

3/2 +
`c_(2)` = 1

`c_(2)` = 1 - 3/2

`c_(2)` = - 1/2

y =
`(3)/(2) e^(x) - (1)/(2) e^(-x)`

12)

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

`c_(1)` +
`c_(2)` = 0 (y(0) = 1) (3)

`c_(1)` = -
`c_(2)`

`e=c_(1) e -c_(2) e^(-1)` (y'(0) = 2) (4)

`e=c_(1) e -(c_(2) )/(e) }`

`e =(c_(1) e^(2) -c_(2) )/(e) }`

`e^(2) ={c_(1) e^(2) -c_(2) }`

replace
`c_(2)` =
`c_(1)` by equation 3

`e^(2) ={c_(1) e^(2) -c_(1) }`

taking common
`c_(1)`

`e^(2) =c_(1) ({e^(2) -1 })`

`\frac{e^(2) }{({e^(2) -1 })} =c_(1)`

`-\frac{e^(2) }{({e^(2) -1 })} =c_(2)`

∴ y =
`(e^(2) )/(1+e^(2) ) (e^(x) - e^(-x) )`

13)

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

`c_(1)`
`e^(-1)` +
`c_(2)`
`e^(1)` = 5 (y(-1) = 5 ) (5)

`c_(1)`
`e^(-1)` -
`c_(2)`
`e^(1)` = -5 (y'(-1) = -5) (6)

Adding 5&6

2
`c_(1)`
`e^(-1)` = 0

`c_(1)` = 0

`c_(2)`
`e^(1)` = 5 -
`c_(1)`
`e^(-1)`

`c_(2)`
`e^(1)` = 5 - 0

`c_(2)`= 5/e

y =
`5e^(-1) e^(-x)`

y =
`5e^(-(x+1))`

14)

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

`c_(1)` +
`c_(2)` = 0 (y(0) = 0) (7)

`c_(1)` -
`c_(2)` = 0 (y'(0) = 0) (8)

Adding 7 & 8

2
`c_(1)` = 0

`c_(2)` =

y = 0