Question

Find the DE:
Equation attached

Answer 1

\[y = C _{1}e ^{3x}+C _{2}e ^{^{-4x}}\]
My answer is \[y'' + y' -12y = 0\]
Just verify me if you found something wrong.

Answer 2

r^2+r-12 = 0
change the y into the r s

Answer 3

and then we factor....
(r+4)(r-3) = 0
r = -4, r = 3

Answer 4

that's about right.

Answer 5

Is that an easy way? What I did is isolation of arbitrary constants.

Answer 6

you could just grab the numbers from that real root equation and do this
(r-3)(r+4)

Answer 7

And then what? I actually have no idea of what you're saying. I was taught by different way.

Answer 8

o-o

Answer 9

Just tell me what to do next. I will do, swear.

Answer 10

I want to know your method so I can compare which one will yield to an answer quicker.

Answer 11

I am in a differential equations course, but my prof hasn't reached that section. I did read about it before school started however

Answer 12

Oww. Okay. ;D

Answer 13

The coefficients of the x in the exponents of the e's tell you all you need to know ^.^

Answer 14

But it seems miss @UsukiDoll already has that covered :D

Answer 15

So what will happen after that?

Answer 16

omg it's PeterPan!!!!

Answer 17

hi hi

Answer 18

They are -4 and 3.
So you need that (minimal) polynomial that has 3 and -4 for its roots.
Namely, (p+4)(p-3) = p^2 + p - 12

Answer 19

doesn't this already look eerily similar to
\[\Large y'' + y'- 12y = 0\]

Answer 20

Am I wrong? Can you tell me the next steps?

Answer 21

PeterPan I need a miracle on my question :D

Answer 22

huh

Answer 23

Differential operator...
\[\Large D \iff \frac{d}{dx}\]

Answer 24

Replace the p in p^2 + p - 12 with the differential operator...
And let the resulting expression be your new differential operator.
\[\Large A:=\left(\frac{d}{dx}\right)^2 + \frac{d}{dx}+12\]

Answer 25

typo alert ^.^
\[\Large A:=\left(\frac{d}{dx}\right)^2 + \frac{d}{dx}-12\]

Answer 26

And your solution will simply be
\[\Large Ay = 0\]

Answer 27

Meaning, of course...
\[\Large A:=\left[\left(\frac{d}{dx}\right)^2 + \frac{d}{dx}-12\right]y=0\]

Answer 28

Do I get the same answer?

Answer 29

Of course you do XD
\[\Large \left.\frac{d^2}{dx^2}y + \frac{d}{dx}y-12y\right.=0\]

Answer 30

\[\Large y'' + y' - 12y = 0\]

Answer 31

Whoa! Welll, great! I thought your explaining that my answer is wrong. LOL XD

Answer 32

I don't know WHAT you did but it sure sounds much more harder than this XD

Answer 33

Hey. Hey. Where do you find the p? and the p^2 and p -12? is that the product of the p-3 and p+4

Answer 34

oh yeah :D
Since the coefficients of the x exponents in the e's are 3 and -4, we need the minimal polynomial having both 3 and -4 for its roots, so obviously, (p-3)(p+4)

p for peter ^.^

Answer 35

Oh I see. hahaha :))
How can be that applicable to other problems?

Answer 36

do you have another problem?

Answer 37

Yes. I already had solutions and answers. Can we apply same thing? I really want fast solutions! \m/

Answer 38

let me see

Answer 39

\[y = c _{1}+c _{2}x+c _{3}e ^{-3x}\]
My answer is just \[9e ^{-3x}d^3y\]

Answer 40

see, this is why i can't answer your 'applicable to other problems' question so directly, as variations can be pretty annoying :>

Answer 41

Well, can you tell me where are that technique be applied.
Btw, I used determinant method to answer that. :D

Answer 42

never heard of it :D

Answer 43

Do you know the correct answer?

Answer 44

No. HAHA

Answer 45

Let's see...

Answer 46

none of this will make sense unless you can do this in reverse, that is, given the differential equation, work out the general solution...

Answer 47

Okay. Say we have y' + y'' -12y = 0

Answer 48

yup... so to do this, find its characteristic equation.

Answer 49

what you do is think of the primes as 'exponents'
that means no prime is a zero-exponent.

so you get
y + y^2 - 12 = 0
and then just solve for y.

it will be confusing if you still use y, so replace it with another letter, say , m.

Answer 50

\[\Large m^2 + m - 12 = 0\]\[\large m = 3 \qquad m = -4\]

Answer 51

So your general solution would be of the form
\[\Large C_1e^{\color{red}3x}+ C_2e^{\color{red}{-4}x}\]

Answer 52

Thanks! I already got the technique. So, does it mean is always involves e?

Answer 53

More or less.
There's a little twist when you get complex roots, but more on that when you actually have to deal with them XD

Answer 54

And there's ANOTHER twist if your roots have multiplicity.

Answer 55

Can you show some? :DD

Answer 56

oh, what the heck
try doing this
\[\large y'' + 6y' + 9y=0\]

Answer 57

\[a e^{-3x} = 0\]Is that the answer? :#

Answer 58

thtat's AN answer.
but it doesn't really cover all possible answers... does it? :D

Answer 59

try \[\Large xe^{-3x}\]

Answer 60

You'll see it fits :D

Answer 61

But I can't form the DE using my method. :/

Answer 62

Okay, chill. When you have roots of multiplicities, in this case, we have -3 twice, then the general solution (at least, the bit involving -3) is as follows...
\[\Large C_1e^{-3x}+C_2\color{red}xe^{-3x}\]

Answer 63

in general, if we have a root r, of multiplicity n, then, the general solution (involving that r) would be
\[\Large C_1e^{rx}+C_2xe^{rx}+...C_nx^{n-1}e^{rx}\]

Answer 64

Okay. I see. I see. I was really having the second though if do I have to put -3x twice. Now I was enlighten. :))
How can I know where to put the variable? Say on our example, the variable x?

Answer 65

what do you mean

Answer 66

Okay. I already see your last reply. :))

Answer 67

good. now back to this problem...

Answer 68

\[\Large y = c _{1}+c _{2}x+c _{3}e ^{-3x}\]

Answer 69

In this part, we see an x-factor (no pun intended)
\[\Large y = \color{green}{c _{1}+c _{2}x}+c _{3}e ^{-3x}\]
and its highest degree is 1, indicating that there is a root of multiplicity 2 :)

Answer 70

But what is that root? :D

Answer 71

Remember how we found the roots? By looking at the coefficient of the x in the exponent of the e, right?

Answer 72

Yes. There is only one root.

Answer 73

There are two. One of them better at hiding than the other ^.^

Answer 74

Where is it?? O.O

Answer 75

Let's hunt for it ^.^

Answer 76

But nothing can hide from me for long :D
\[\Large y = c _{1}\color{red}{e^{\color{black}0x}}+c _{2}x\color{red}{e^{\color{black}0x}}+c _{3}e ^{-3x}\]

NOW do you understand? :D

Answer 77

Yes. I know that.

Answer 78

Okay, so the roots are -3 and 0.
Except 0 is taken as a root TWICE since there's an x-factor. (note that if there was an x^2 factor, then that means that 0 is taken as a root THREE times, etc)

Answer 79

Is it (t+3)^3 ??

Answer 80

No.

Answer 81

Roots are 0, 0, and -3.
So...
(p - 0)(p - 0)(p + 3)

Answer 82

p^2 (p+3)

Answer 83

(p^3+3p^2)

Answer 84

Yup ^.^

Answer 85

So, that means...?

Answer 86

Is it y''' + 3y'' = 0 ?

Answer 87

that's right ^.^

Answer 88

Owww. Thank you. I got new lessons from you. :))

Answer 89

:)

Answer 90

But I found different answer. :/ We're actually forced to do determinant method in this question, but don't worry. I'll just recheck it.