Question

DE with e^x

Answer 1

@ganeshie8 @Empty

Answer 2

\[e^xy''+xy=0\]

Answer 3

baby
\[\int e^x\]
;)

Answer 4

\[e^x \sum_{n=2}^{\infty} a_n n(n-1)x^{(n-2)}+x \sum_{n=0}^{\infty} a_n x^n = 0\] lets note \[e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! }\]

Answer 5

so power series

Answer 6

ye

Answer 7

is it mandatory to use that

Answer 8

Yeah haha

Answer 9

\[\sum_{n=0}^{\infty} \frac{ x^n }{ n! } \sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)x^n+\sum_{n=0}^{\infty} a_n x^{n+1}\]

\[\left( 1+\frac{ x }{ 1! }+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }+... \right)\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)x^n+\sum_{n=0}^{\infty} a_nx^{n+1}\]

Answer 10

I guess we need x^(n+1) in the first sum or something, maybe I did it wrong haha.

Answer 11

Oooh wait a sec

Answer 12

Maybe if \[\left( 1+x+\frac{ x^2 }{ 2 }+\frac{ x^3 }{ 3! }+... \right)\left( 2a_2+3a_3x+... \right)+\sum_{n=0}^{\infty} a_n x^{n+1}=0\]

Answer 13

Not sure xD

Answer 14

I have to find four nonzero terms in both the power series solutions about the origin, maybe I should have wrote that to, might help...

Answer 15

simply set the coefficients equal to 0

Answer 16

\[\left( 1+x+\frac{ x^2 }{ 2 }+\frac{ x^3 }{ 3! }+... \right)\left( 2a_2+3a_3x+... \right)+\sum_{n=0}^{\infty} a_n x^{n+1}=0\]

whats the constant term on left hand side ?

Answer 17

1

Answer 18

think again

Answer 19

\[\left( \color{red}{1}+x+\frac{ x^2 }{ 2 }+\frac{ x^3 }{ 3! }+... \right)\left( \color{red}{2a_2}+3a_3x+... \right)+\sum_{n=0}^{\infty} a_n x^{n+1}=0\]


multiplying those two red guys is the only way to get a constant term on left hand side, yes ?

Answer 20

Yeah haha

Answer 21

I was just writing that

Answer 22

since there are no constant terms on right hand side, it must be the case that
\[2a_2=0\tag{1}\]

Answer 23

similarly, try comparing the coefficent of "x" both sides

Answer 24

Ok I get the first one, but would I need to expand the other sum to then?

Answer 25

you may get it by staring at the sum :)

Answer 26

Ok let me see, I think I get it, this technique is pretty neat lol

Answer 27

\[\left( 1+\frac{ x }{ 1! }+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }+... \right)\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)x^n+\sum_{n=0}^{\infty} a_nx^{n+1}\]

comparing constant terms :
\[2a_2=0\tag{1}\]

comparing x coefficients :
\[2a_2+6a_3+a_0=0\tag{2}\]

comparing x^2 coefficients :
\[a_2+6a_3+12a_4+a_1=0\tag{3}\]

Answer 28

solving them gives

\(a_2=0\)

\(a_3=-a_0/6\)

\(a_4=(a_0-a_1)/12\)

Answer 29

just compare x^3 and x^4 coefficients too as you need "four" nonzero terms...

Answer 30

That makes a lot of sense! Thanks a lot ganeshieeeeeeeee!

Answer 31

All of these questions have like 10 parts, but once you figure out how to do this the rest is easy haha.

Answer 32

As you can see, we are expressing the coefficients in terms of \(a_0\) and \(a_1\).

The differential equation is of second order, so the general solution requires exactly "two" arbitrary constants. These are \(a_0\) and \(a_1\) here...

Answer 33

Yup, the solution is \[y_1(x) = 1- 1/6x^3 + 1/12 x^4-1/40x^5\] then \[y_2(x) = x-1/12x^4+1/20x^5-1/60x^6+...\]

Answer 34

oh by the way can I ask you like a theoretical question?

Answer 35

what happened to the arbitrary constants ?

Answer 36

say no say no, not allowed to ask a theoretical question! xD

Answer 37

ohk, you're setting \(a_0=0, a_1=1\) etc... nice :)

Answer 38

Yeah exactly

Answer 39

say yes plsss

Answer 40

these are two independent solutions
\[y_1(x) = 1- 1/6x^3 + 1/12 x^4-1/40x^5\] \[y_2(x) = x-1/12x^4+1/20x^5-1/60x^6+...\]

so we may write the general solutoin as \(y = c_1y_1 + c_2y_2\)

Answer 41

Yup! I know how to do the rest I guess, it's just wasn't sure on the approach when given e^x

Answer 42

I think, saying "\(\cdots \)" is the general solution is meaningless
so we don't really know the general solution yet.... we just know the first four terms i guess..

Answer 43

I think you need the general term probably for the general solution xD

Answer 44

exactly, i could say
\[\pi = 3.142\ldots\]

which is simply nonsense because that doesn't tell you how to get the subsequent digits...

Answer 45

I was just thinking of this exact same example hahaha, that's why I have a lot of beef with "elementary" functions and so-called "closed-form" solutions. They simply don't exist like people like to believe.

Answer 46

i think below is meaningful to say :
\[\dfrac{1}{3}=0.333\ldots = 0.\overline{3}\]

Answer 47

because, the reader would know what it is then..

Answer 48

I agree xD

Answer 49

In some sense, it might be easier to say in base 3, we express a third as \(.1\) haha

Answer 50

I don't even know how that works

Answer 51

I should take analysis, you learn a lot of this stuff in it...even induction, which I've actually never really used

Answer 52

not like faraday but math induction

Answer 53

its easy, and you would have known it from childhood if you're yoda with 3 fingers.. then you will be forced to use base 3 because you can only count upto 3 objects using your fingers.. :) jk..

Answer 54

Ahahaha, guess we gotta chop off 7 of your fingers now.

I kinda dislike analysis and induction. Induction can really only confirm something is true but generally it sorta misses the point. Induction is useless if you derived the expression already.

Answer 55

Isn't induction what we essentially use here when we get like a recurrence relation

Answer 56

I can see how induction and analysis must be hated together...

Induction doesn't help in getting the formula, but it is still a beatiful proof mehtod because it confirms the formula you have got works in "general".

It doesn't help in getting the intuitoin of why the formula works, but it is powerful because it tells you immediately if your intuition is correct or wrong..

Answer 57

I noticed something funny here, I want to work this out:

\[y'' = -xe^{-x} y\]
we can write \(y''\) in two ways using the chain rule:

\[\frac{d (y')}{dx} = \frac{d (y')}{dy}\frac{dy}{dx} = y' \frac{d(y')}{dy}\]

So now we have two similar integrals we can solve... But in both cases we've left out the other variable. Hmm

\[\int dy' = \int -xe^{-x} y dx\]
and
\[\int y' dy' = \int -xe^{-x} y dy\]

Answer 58

I never understood why we couldn't just do that from the beginning

Answer 59

Well we can, and I just did, but it doesn't lead us anywhere since we're stuck. These integrals won't work since we can only integrate over one variable x or y.

Answer 60

I really want to just take these and multiply them together to integrate it like this:

\[\tfrac{1}{2}(y')^3 = \iint -xe^{-x}y dA\]