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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

Respuesta :

Answer:

11)y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)y = [tex]5e^{-(x+1)}[/tex]

14)y = 0

Step-by-step explanation:

Given data:

[tex]y=c_{1} e^{x} +c_{2} e^{-x}[/tex]

y''-y=0

The equation is

[tex]m^{r}[/tex]-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

[tex]y'=c_{1} e^{x} -c_{2} e^{-x}[/tex]

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 1   (y(0) = 1) (1)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 2   (y'(0) = 2)  (2)

adding 1 & 2

2[tex]c_{1}[/tex] = 3

[tex]c_{1}[/tex] = 3/2

3/2 +  [tex]c_{2}[/tex] = 1

[tex]c_{2}[/tex]  = 1 -  3/2

[tex]c_{2}[/tex] = - 1/2

y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)

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

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 0 (y(0) = 1) (3)

[tex]c_{1}[/tex] = - [tex]c_{2}[/tex]

[tex]e=c_{1} e -c_{2} e^{-1}[/tex]   (y'(0) = 2)  (4)

[tex]e=c_{1} e -\frac{c_{2} }{e} }[/tex]

[tex]e =\frac{c_{1} e^{2} -c_{2} }{e} }[/tex]

[tex]e^{2} ={c_{1} e^{2} -c_{2} }[/tex]

replace [tex]c_{2}[/tex] = [tex]c_{1}[/tex] by equation 3

[tex]e^{2} ={c_{1} e^{2} -c_{1} }[/tex]

taking common [tex]c_{1}[/tex]

[tex]e^{2} =c_{1} ({e^{2} -1 })[/tex]

[tex]\frac{e^{2} }{({e^{2} -1 })} =c_{1}[/tex]

[tex]-\frac{e^{2} }{({e^{2} -1 })} =c_{2}[/tex]

∴ y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)

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

[tex]c_{1}[/tex][tex]e^{-1}[/tex] +  [tex]c_{2}[/tex][tex]e^{1}[/tex] = 5   (y(-1) = 5 ) (5)

[tex]c_{1}[/tex][tex]e^{-1}[/tex] -  [tex]c_{2}[/tex][tex]e^{1}[/tex] = -5    (y'(-1) = -5)  (6)

Adding 5&6

2[tex]c_{1}[/tex] [tex]e^{-1}[/tex] = 0

[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - [tex]c_{1}[/tex][tex]e^{-1}[/tex]

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - 0

[tex]c_{2}[/tex]= 5/e

y = [tex]5e^{-1} e^{-x}[/tex]

y = [tex]5e^{-(x+1)}[/tex]

14)

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

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] =  0 (y(0) = 0) (7)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 0   (y'(0) = 0)  (8)

Adding 7 & 8

2[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex] =

y = 0

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