- anonymous

(D^2 + 9)y =5e^x - 162x
i need to check if my answer is right.

- katieb

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- anonymous

@AnimalAin this.

- anonymous

Awesome! u did it write

- anonymous

this is my answer:
2Ae^x + B + Cx = 5e^x -162x

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## More answers

- anonymous

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

- anonymous

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.

- anonymous

lilmissmindset is right thats what i got for my answer

- anonymous

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

- anonymous

yes yes, this is the first question i asked. it's just that the book says the general solution for this is
\[=C _{1} \cos 3x+ C _{2} \sin3x + \frac{ 1 }{ 2 } e ^{x} -18xy \]

- anonymous

there's no y after -18x SORRY

- anonymous

Lets look at this in more detail.

- anonymous

and our answer is
5/2 e^x -162x

- anonymous

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

- anonymous

\[(D^2 + 9)y =5e^x - 162x\]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?

- anonymous

yeps. ok.

- anonymous

ok with that part,

- anonymous

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

- anonymous

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.

- anonymous

owwww. i understand,

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

not*

- anonymous

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

- anonymous

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

- anonymous

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,

- anonymous

SORRY!

- anonymous

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

- anonymous

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

- anonymous

i'll practice harder, :)

- anonymous

I'm sure you'll get it.

- anonymous

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

- anonymous

may i ask what book the problem referred?

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