## Trending News

# y′′′−2y′′−y′+2y=0, y(0)=−8, y′(0)=2, y′′(0)=−11?

find y(x). I did this multiple times but keep getting it wrong.

### 2 Answers

- la consoleLv 72 years agoFavorite Answer
y′′′ - 2y′′ - y′ + 2y = 0

Characteristic equation:

r³ - 2r² - r + 2 = 0

(r³ - 2r²) - (r - 2) = 0

r².(r - 2) - (r - 2) = 0

(r² - 1).(r - 2) = 0

(r + 1).(r - 1).(r - 2) = 0 → the roots are: - 1 ; 1 ; 2

y = C₁.e^(- x) + C₂.e^(x) + C₃.e^(2x) ← this is the general solution → then you can deduce that:

y’ = - C₁.e^(- x) + C₂.e^(x) + 2.C₃.e^(2x)

y’’ = C₁.e^(- x) + C₂.e^(x) + 4.C₃.e^(2x)

y’’’ = - C₁.e^(- x) + C₂.e^(x) + 8.C₃.e^(2x)

First condition: y(0) = - 8

y = C₁.e^(- x) + C₂.e^(x) + C₃.e^(2x) → when: x = 0, the result is - 8

C₁.e^(0) + C₂.e^(0) + C₃.e^(0) = - 8 → you know that: x^(0) = 1 whatever the value of x

C₁ + C₂ + C₃ = - 8

C₃ = - 8 - C₁ - C₂ ← equation (1)

Second condition: y’(0) = 2

y’ = - C₁.e^(- x) + C₂.e^(x) + 2.C₃.e^(2x) → when: x = 0, the result is 2

- C₁.e^(0) + C₂.e^(0) + 2.C₃.e^(0) = 2 → recall: x^(0) = 1

- C₁ + C₂ + 2.C₃ = 2 → recall (1): C₃ = - 8 - C₁ - C₂

- C₁ + C₂ + 2.(- 8 - C₁ - C₂) = 2

- C₁ + C₂ - 16 - 2.C₁ - 2.C₂ = 2

- 3.C₁ - C₂ = 18

C₂ = - 3.C₁ - 18 ← equation (2)

Third condition: y’’(0) = - 11

y’’ = C₁.e^(- x) + C₂.e^(x) + 4.C₃.e^(2x) → when: x = 0, the result is - 11

C₁.e^(0) + C₂.e^(0) + 4.C₃.e^(0) = - 11 → recall: x^(0) = 1

C₁ + C₂ + 4.C₃ = - 11 → recall (1): C₃ = - 8 - C₁ - C₂

C₁ + C₂ + 4.(- 8 - C₁ - C₂) = - 11

C₁ + C₂ - 32 - 4.C₁ - 4.C₂ = - 11

- 3.C₁ - 3.C₂ = 21

C₁ + C₂ = - 7 → recall (2): C₂ = - 3.C₁ - 18

C₁ + (- 3.C₁ - 18) = - 7

C₁ - 3.C₁ - 18 = - 7

- 2.C₁ = 11

→ C₁ = - 11/2

Recall (2): C₂ = - 3.C₁ - 18

C₂ = - 3.(- 11/2) - 18

C₂ = (33/2) - (36/2)

→ C₂ = - 3/2

Recall (1): C₃ = - 8 - C₁ - C₂

C₃ = - 8 - (- 11/2) - (- 3/2)

C₃ = - 8 + (11/2) + (3/2)

→ C₃ = - 1

Recall the gneral solution:

y = C₁.e^(- x) + C₂.e^(x) + C₃.e^(2x) → where: C₁ = - 11/2 → where: C₂ = - 3/2 → where: C₃ = - 1

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

→ y = - [11.e^(- x) + 3.e^(x) + 2.e^(2x)] / 2

- VamanLv 72 years ago
Use y=exp ax

Put this in the equation, You get

a³ -2a²-a+2=0a² (a-2)-(a-2)=0. (a-2) (a²-1)=0. This gives three roots 2 , +1 and -1. We expect solution of the form

y = bexp 2x + c exp-x+d exp x. putting back in the solution y=exp ax.

You need to find b , c and d

t=0 y(0)=-8 This gives -8= b+c+d (1)

t=0 y'(0) = 2 This gives 2= 2b +c-d (2)

t-0 y''(0) = -11. this gives -11= 4b +c+d (3)

Find b, c and d. You get the solution

y(x) = -13/2 exp 2x +27/4 exp x -33/4 exp -x