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

Answer:

11)y = $\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}$

12)y = $\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )$

13)y = $5e^{-(x+1)}$

14)y = 0

Step-by-step 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 = $\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}$

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)

$e=c_{1} e -\frac{c_{2} }{e} }$

$e =\frac{c_{1} e^{2} -c_{2} }{e} }$

$e^{2} ={c_{1} e^{2} -c_{2} }$

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 = $\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )$

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