anonymous
  • anonymous
(D^2 + 9)y =5e^x - 162x i need to check if my answer is right.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@AnimalAin this.
anonymous
  • anonymous
Awesome! u did it write
anonymous
  • anonymous
this is my answer: 2Ae^x + B + Cx = 5e^x -162x

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anonymous
  • anonymous
lets see...... AWESOME AWESOME :) thats right good job
anonymous
  • 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
  • anonymous
lilmissmindset is right thats what i got for my answer
anonymous
  • anonymous
That's the problem we did before.....
anonymous
  • 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
  • anonymous
there's no y after -18x SORRY
anonymous
  • anonymous
Lets look at this in more detail.
anonymous
  • anonymous
and our answer is 5/2 e^x -162x
anonymous
  • anonymous
i thought, maybe there's something wrong with me, getting the solution after all
anonymous
  • 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
  • anonymous
yeps. ok.
anonymous
  • anonymous
ok with that part,
anonymous
  • 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
  • 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
  • anonymous
owwww. i understand,
anonymous
  • 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
  • anonymous
So we get something like\[y=y_h+y_p=C_1\cos 3x+C_2\sin3x+\frac{e^x}{2}-19x\]
anonymous
  • anonymous
i forgot to multiply \[y _{p}\] with 9 when i substitute, right? that's why i did ot get the answer,
anonymous
  • anonymous
not*
anonymous
  • anonymous
Actually, my division is horrible today.... should be 18 in the last term; exactly as your text had said.
anonymous
  • anonymous
I hope you've got this; you're wearing me out....LOL
anonymous
  • 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
  • anonymous
SORRY!
anonymous
  • anonymous
i get it, i'm just confused on some parts.
anonymous
  • anonymous
No stress. It's fun working with students that work hard and get it. Do math every day.
anonymous
  • anonymous
i'll practice harder, :)
anonymous
  • anonymous
I'm sure you'll get it.
anonymous
  • anonymous
i hope i get it the soonest possible, my exam is tomorrow. O.O
anonymous
  • anonymous
may i ask what book the problem referred?

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