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.

(1 pt) We consider the non-homogeneous problem y-16y= 256x^2 First we consider the homogeneous problem y-16y=0: 1) the auxiliary equation is ar^2 + br + c =r^2-16=0 2) The roots of the auxiliary equation are 4, -4 (enter answers as a comma separated list). 3) A fundamental set of solutions is e^(4x), e^(-4x) (enter answers as a comma separated list). Using these we obtain the the complementary solution yc= c1y1 + c2y2 for arbitrary Constants c1 and c2 Next we seek a particular solution yp of the non-homogeneous problem y-16y-256 x^2 using the method of undetermined coefficients (See the link below for a help sheet) 4) Apply the method of undetermined coefficients to find yp=-16x^2 We then find the general solution as a sum of the complementary solution yc= c1y1 +c2y2 and a particular solution: y =yc +yp. Finally you are asked to use the general solution to solve an IVP. 5) Given the initial conditions y(0)=-3 and y?(0)=-4 find the unique solution to the IVP Y= 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.

Explanation / Answer

On the right, you have 256x^2, which is a quadratic

So, the particular solution is the standard quadratic....

yp = ax^2 + bx + c

So, yp' = 2ax + b

yp'' = 2a

Now, we had y'' - 16y = 256x^2

So, 2a - 16(ax^2 + bx + c) = 256x^2

(-16a)x^2 + (-16b)x + (2a - 16c) = 256x^2

Equating x^2 terms ---> -16a = 256 --> a = -16

Equating x trms ---> -16b = 0 --> b = 0

Equating constants --> 2a - 16c = 0 --> -32 - 16c = 0 --> c = -2

So, y'(x) = ax^2 + bx + c ---> yp(x) = -16x^2 + 0x + (-2)

yp = -16x^2 - 2 ---> FOURTH ANSWER

So, y = yc + yp

y = c1e^(4x) + c2e^(-4x) + (-16x^2 - 2)

y = c1e^(4x) + c2e^(-4x) - 16x^2 - 2

y(0) = -3 :
-3 = c1 + c2 - 2
c1 + c2 = -1 ---> ONE

y' = 4c1e^(4x) - 4c2e^(-4x) - 32x
y'(0) = -4 :
-4 = 4c1 - 4c2
c1 - c2 = -1 ---> TWO

From ONE and TWO :

c1 + c2 = -1
c1 - c2 = -1

Adding :
2c1 = -2
c1 = -1 and c2 = 0

y = c1e^(4x) + c2e^(-4x) - 16x^2 - 2

which now becomes :

y = -e^(4x) - 16x^2 - 2 -----> FIFTH ANSWER

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