PLEASE SHOW ALL WORK for these last two parts. I fully understand the parts I ha

Question

PLEASE SHOW ALL WORK for these last two parts. I fully understand the parts I have gotten correct. It is these last two parts - mainly the solving for y(p) - that is throwing me for a loop. Also if you could give a breif comment on how exactly I could apply it to find y(p) on other problems.

Courses Homework Sets HW8: Problem 2 HW8 Problem 2 PasswordEmail Prev Up Next Grades Entered Answer Preview Result Problems (rA2)-6 T+9 correct Problem 1 Problem 2 Problem 3 Problem4 Problem 5 Problem 6 Problem 7 Problem 8 Problem9 Problem 10 3, 3 correct 3,3 e*(3%), x"e"(3%) correct 3z incorrect incorrect At least one of the answers above is NOT correct 2 of the questions remain unanswered (1 pt) We consider the non-homogeneous problem y"-6y' 9y Display Options 81r3 View equations as o images o MathJax Show saved answers? First we consider the homogeneous problem y"-6y' 9y 0 1) the auxiliary equation is ar2 + br crA2-6r+9 2) The roots of the auxiliary equation are 3,3 3) A fundamental set of solutions is e(3x), xeA (3x) 0 Yes (enter answers as a comma separated list) No Apply Options (enter answers as a comma separated list). Using these we obtain the the complementary solution ye C1y1 + c2y2 for arbitrary constants ci and C2 Next we seek a particular solution yp of the non-homogeneous problem y"- 6y' 9y 81z3 using the method of undetermined coefficients (See the link below for a help sheet) 4) Apply the method of undetermined coefficients to find Yp- We then find the general solution as a sum of the complementary solution ye- C1y1 IVP c2U2 and a particular solution: y yc +yp Finally you are asked to use the general solution to solve an 5) Given the initial conditions y(0) 9 and y'(0) 24 find the unique solution to the IVP Help Sheet: Undetermined Coefficients Notes Note: You can earn partial credit on this problem

Explanation / Answer

On the right, we have 81x^3, so the particular solution is of the form :

yp = ax^3 + bx^2 + cx + d

yp' = 3ax^2 + 2bx + c

yp'' = 6ax + 2b

So, y'' - 6y' + 9y = 81x^3

becomes

6ax + 2b - 6( 3ax^2 + 2bx + c) + 9( ax^3 + bx^2 + cx + d) = 81x^3

6ax + 2b - 18ax^2 - 12bx - 6c + 9ax^3 + 9bx^2 + 9cx + 9d = 81x^3

(9a)x^3 + (-18a + 9b)x^2 + (6a - 12b + 9c)x + (2b - 6c + 9d) = 81x^3

Comparing x^3 --> 9a = 81 ---> a = 81/9 ---> a = 9

Comparing x^2 ---> -18a + 9b = 0 --> 2a = b --> b = 2(9) ---> b = 18

Comparing x ----> 6a - 12b + 9c = 0 ----> 54 - 216 + 9c = 0 --> 9c = 162 --> c = 18

Comparing constants --> 2b - 6c + 9d = 0 ---> 36 - 108 + 9d = 0 ---> 9d = 72 ---> d = 8

So, ax^3 + bx^2 + cx + d becomes :

yp = 9x^3 + 18x^2 + 18x + 8 ---> FIRST ANSWER

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So, y = yc + yp becomes

y = c1*e^(3x) + c2*x*e^(3x) + 9x^3 + 18x^2 + 18x + 8

y(0) = 9 :

9 = c1 + 0 + 0 + 0 + 0 + 8

c1 = 9 - 8

c1 = 1

y'(0) = 24

Lets find y' first

y' = 3c1e^(3x) + c2e^(3x) + 3c2*x*e^(3x) + 27x^2 + 36x + 18

24 = 3*1 + c2 + 0 + 0 + 0 +18

24 = 21 + c2

c2 = 3

So, final solution becomes :

y =1*e^(3x) + 3*x*e^(3x) + 9x^3 + 18x^2 + 18x + 8 ---> ANSWER

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