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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Awesome! u did it write

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is my answer: 2Ae^x + B + Cx = 5e^x 162x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lets see...... AWESOME AWESOME :) thats right good job

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It'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
 3 years ago
Best ResponseYou've already chosen the best response.0lilmissmindset is right thats what i got for my answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's the problem we did before.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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
 3 years ago
Best ResponseYou've already chosen the best response.0there's no y after 18x SORRY

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lets look at this in more detail.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and our answer is 5/2 e^x 162x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought, maybe there's something wrong with me, getting the solution after all

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(D^2 + 9)y =5e^x  162x\]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
 3 years ago
Best ResponseYou've already chosen the best response.0Then 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
 3 years ago
Best ResponseYou've already chosen the best response.0This 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now consider the polynomial part. The second derivative of a linear function is zero, so we get\[ 9y_{p2} = 162x \implies y_{p2} =19x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So we get something like\[y=y_h+y_p=C_1\cos 3x+C_2\sin3x+\frac{e^x}{2}19x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i forgot to multiply \[y _{p}\] with 9 when i substitute, right? that's why i did ot get the answer,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, my division is horrible today.... should be 18 in the last term; exactly as your text had said.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I hope you've got this; you're wearing me out....LOL

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually, 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
 3 years ago
Best ResponseYou've already chosen the best response.0i get it, i'm just confused on some parts.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No stress. It's fun working with students that work hard and get it. Do math every day.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'll practice harder, :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sure you'll get it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i hope i get it the soonest possible, my exam is tomorrow. O.O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0may i ask what book the problem referred?
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