Question

(D^2 + 9)y =5e^x - 162x

i need to check if my answer is right.

Answer 1

@AnimalAin this.

Answer 2

Awesome! u did it write

Answer 3

this is my answer:

2Ae^x + B + Cx = 5e^x -162x

Answer 4

lets see...... AWESOME AWESOME :) thats right good job

Answer 5

It's been a minute since I've done these, but I think the way is to consider the homogeneous solution by using an auxilliary (sp?) equation, then fit a particular solution that will match the RHS.

Answer 6

lilmissmindset is right thats what i got for my answer

Answer 7

That's the problem we did before.....

Answer 8

yes yes, this is the first question i asked. it's just that the book says the general solution for this is

\[=C _{1} \cos 3x+ C _{2} \sin3x + \frac{ 1 }{ 2 } e ^{x} -18xy \]

Answer 9

there's no y after -18x SORRY

Answer 10

Lets look at this in more detail.

Answer 11

and our answer is

5/2 e^x -162x

Answer 12

i thought, maybe there's something wrong with me, getting the solution after all

Answer 13

\[(D^2 + 9)y =5e^x - 162x\]Consider\[(D^2 + 9)y =0\]Aux equation\[r^2+9=0 \implies r= \pm 3i \implies y _{h}=C _{1} \cos 3x +C_2 \sin3x\]OK so far?

Answer 14

yeps. ok.

Answer 15

ok with that part,

Answer 16

Then we find the particular solution. Consider the first part of the RHS by itself, so\[(D^2 + 9)y_p =5e^x \implies y_p=\frac{e^x}{2}\]

Answer 17

This is true since the derivative of the e^x term is always itself. That means the second derivative is the same as the original function. Added to the nine copies of the original function, we get 10ysubp=5e^x.

Answer 18

owwww. i understand,

Answer 19

Now consider the polynomial part. The second derivative of a linear function is zero, so we get\[ 9y_{p2} =- 162x \implies y_{p2} =-19x\]

Answer 20

So we get something like\[y=y_h+y_p=C_1\cos 3x+C_2\sin3x+\frac{e^x}{2}-19x\]

Answer 21

i forgot to multiply \[y _{p}\] with 9 when i substitute, right? that's why i did ot get the answer,

Answer 22

not*

Answer 23

Actually, my division is horrible today.... should be 18 in the last term; exactly as your text had said.

Answer 24

I hope you've got this; you're wearing me out....LOL

Answer 25

actually, it's the same for the two of us. LOL

hey, thanks a lot. you've been helping me large. Sorry for the stress,

Answer 26

SORRY!

Answer 27

i get it, i'm just confused on some parts.

Answer 28

No stress. It's fun working with students that work hard and get it. Do math every day.

Answer 29

i'll practice harder, :)

Answer 30

I'm sure you'll get it.

Answer 31

i hope i get it the soonest possible, my exam is tomorrow. O.O

Answer 32

may i ask what book the problem referred?