Could someone help me with this problem D WeBWorK BaerMAT2

Could someone help me with this problem?

D WeBWorK : Baer-MAT-2 >New Tab -> c https://webwork2.asu.edu/webwork2/Baer-MAT.275-Fall-2016/Section_3.4. Bookmarks ts/8/?key-GOhJzLuopAJ83nARV1TRvzd7EHXz5zW&user-jnbyun;&effectiveUser-jnbyun; Other bookmarks ( WeBWorK Logged in as jnbyun Log OutG MATHEMATICAL ASSOCIATION OF AMERICA webwork / baer mat 275-fall-2016 / section-3.4-repeated-roots/ 8 MAIN MENU Courses Homework Sets Section 3.4 Repeated Roots Section 3.4 Repeated Roots: Problem 8 Problem 8 Previous Problem List Next User Settings Grades (1 point) Find y as a function of z if Problems y(0)=-6, 3/(0)= 3, y\"(0)= 99 Problem 1 Problem 2V Problem 3. Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Email instructor 12:06

Solution

Let,

y\"\' - 6 y\" - y\' +6y = 0

with y(0) = -6 , y\'(0) = 3 , y\" (0) =99

The characteristic equation and its roots are,

r³ - 6r² - r +6 = 0

r² ( r - 6) - 1 (r - 6) = 0

( r² -1) (r-6) = 0

( r -1) ( r+1) ( r- 6) = 0

therefore r = 1 , r = 6 , r = -1

the genral solution of its derivative are

y( x ) = C₁ e^x + C₂ e^6x + C₃e^-x   ........................................1

y\' ( x) = C₁e^x +6 C₂ e^6x - C₃ e^-x ........................................ 2

y\" (x) = C₁ e^x +36 C₂ e^6x + C₃ e^-x ..........................................3

putting in the initial conndition in the above equation

Gives the following system of equation

for y(0) = -6 , to 1

-6 = C₁ e⁰ + C₂ e⁰ + C₃ e ⁰

C₁ + C₂ + C₃  = -6 ...............................................................4

for y\'(0) = 3 , to 2

C₁ + 6 C₂  - C₃ = 3 .................................................................5

for y\" (0 ) = 99, to 3

C₁ + 36 C₂ + C₃ = 99 ..............................................................6

Adding 4 and 5 we get

2C₁ + 7 C₂  = -3 ..........................................................................7

Adding 5 and 6 we get

2C₁ + 42 C₂  = 102 .......................................................................8

Solving 7 and 8 we get

substracting 7 from 8 we get

C₁ = 9 amd C₂ = -3

puttind  C₁ = 9 amd C₂ = -3 in 4 we get

9 - 3 + C₃ = 6

C₃ = 0

Therefore the solution is

y(x) = 9 e^x - 3 e^6x + 0 e^-x

= 9 e^x - 3 e^6x

= 3 (  3 e^x - e^6x )

Done