Question

How did I do?
2nd order DE
\[25y''+9y=0\]
\[r(r+9)=0\]
so r=0 and r=-9
\[y=c_2e^{-9x}\]

Answer 1

im confused on what you are asking

Answer 2

The question in my book reads: Solve the differential equation. I think I did that correctly, but let me know if I made any mistakes.

Answer 3

Uhh... yeah, sorry, it should be \[
c_1e^{\frac{3}{5}ix}+c_2e^{-\frac{3}{5}ix}=y(x)\]

Answer 4

why?

Answer 5

Characteristic polynomial (roots of)

Answer 6

duh. I wrote it wrong in my notebook....I solved r^2+9r=0 and it should have been 25r^2+9r

Answer 7

I recommend \(25r^2+9=0\) ... I think you'd benefit from checking your work, again.

Answer 8

yeah. sorry for my dumb little mistake :P

Answer 9

It's all right, it happens.

Answer 10

I still get a different answer than you did.
\[25r^2+9r=0\]
\[r(25r+9)=0\]
so r=0 and \[r=\frac{-9}{25}\]
\[y=c_2e^{\frac{-9}{25}x}\]

Answer 11

@LolWolf ?

Answer 12

Wait, how? We take:
\[
25r^2e^{rx}+9e^{rx}=0
\]And factor \(e^{rx}\) out, which we know cannot be zero, thus:
\[
(25r^2+9)e^{rx}=0 \implies (25r^2+9)=0
\]

Answer 13

(Remember you don't differentiate \(y(x)\) ...! Important)

Answer 14

Is my auxiliary equation wrong?
\[r(25r+9)=0\]

Answer 15

Yeah, I think you're differentiating \(y(x)\) as the polynomial should be \(25r^2+9=0\)

Answer 16

See question 6

Answer 17

Ooooh...I see what I did. the auxiliary equation is
\[r^2+9=0\]

Answer 18

I can't read today LOL :P

Answer 19

Oh gawd!
\[25r^2+9=0\]

Answer 20

Yes, here we have:
\[
25\frac{d^2y(x)}{dx^2}+9y(x)=0 \\
y(x)=e^{rx}\\
25\frac{d^2e^{rx}}{dx^2}+9e^{rx}=0 \implies\\
25r^2e^{rx}+9e^{rx}=0 \implies\\
(25r^2+9)e^{rx}=0 \implies\\
25r^2+9=0
\]Et al, you know the drill.

Answer 21

thank you. Thanks for patient so patient with me. That makes more sense now

Answer 22

I can't write today either. I meant thanks for being soo patient with me. I worked the night shift, my brain is not quite up to speed today. :P

Answer 23

Sure thing, glad to help. And, it's all right.