Question

Solve the IVP x" - 2tx' + 4x = 0, x(0) = -1, x'(0) = 0.
How to start?

Answer 1

what do you know?

Answer 2

the initial condtions bring laplace to mind

Answer 3

variation of parameters comes to mind too

Answer 4

But... L(2tx')?? Not sure how to do it...

Answer 5

sL(gf) = L(g'f) + L(gf') + gofo

might be useful

Answer 6

i dont know all the tabled stuff, so i cant just bring something to the mat

Answer 7

I can't recall I have learnt that formula...

Answer 8

you know L(y') = sL(y) - yo right?

Answer 9

Yup!

Answer 10

by the product rule

L(g'f + gf') = sL(gf) - gofo

by linearity

L(g'f) + L(gf') = sL(gf) - gofo

and by addition

L(g'f) + L(gf') + gofo = sL(gf)

Answer 11

not sure how well it plays out, but i know it plays :)

Answer 12

For variation of parameter, not sure how this question can be solved using that technique since not all coefficients are constants nor do we have the fundamental set of solutions.

Answer 13

sL(t t) = L(t*1) + L(1*t) + 0,0

sL(t^2) = 2L(t)

s^2 L(t^2) = 2s L(t)
-----------------------

s L(t*1) = L(1*1) + L(t*0) + 0*1

s L(t) = L(1)

s^3 L(t^2) = 2 sL(1)
----------------------

s L(1*1) = L(0) + L(0) + 1

s L(1) = 1
--------------------

s^3 L(t^2) = 2

L(t^2) = 2/s^3

Answer 14

2s L(t x') = L(1*x') + L(t x'') + 0*x'(0)

2s^2 L(t x') = sL(x') + sL(t x'')

2s^2 L(t x') = s(sL(x)-x(0)) + sL(t x'')
---------------------------

sL(t x'') = L(x'') + L(t x''')

i see that going in circles, unless we can see a more clever trick to it

the table says:

L(t x') = -dx/ds

Answer 15

Just wondering, would it be possible to use series method here? It is odd though, since we are given the initial conditions....

Answer 16

L(x') = x(s)

L(t x') = -x'(s)

Answer 17

yeah, series is fine

Answer 18

But then we have to check ordinary point etc etc?!

Answer 19

dont see why we would have to ....

Answer 20

We assume the solution being sum from k=0 to infty a_k (t-a)^k, where a = ordinary point?!

Answer 21

assuming my interpretation of a table is right

x" - 2tx' + 4x = 0

s(sL(x)-x(0)) - x'(0) + 4L(x) + 2x'(s) = 0

x(0) = -1, x'(0) = 0

s(sL(x)+1) + 4L(x) + 2x'(s) = 0

s^2 L(x) +s+ 4L(x) + 2x'(s) = 0

(s^2+4) L(x) = - 2x'(s) - s

L(x) = -[2x'(s) + s]/[s^2+4]

Answer 22

Unless I can derive L(t x') = -dx/ds in the exam, I cannot use this result since I have not been taught about this in class...

Answer 23

well, im saying that it might suggest a cos function?

Answer 24

well that was a boring solution lol

Answer 25

How so?!

Answer 26

i peeked at the wolfs output

Answer 27

https://www.wolframalpha.com/input/?i=x%22+-+2tx%27+%2B+4x+%3D+0%2C+x%280%29+%3D+-1%2C+x%27%280%29+%3D+0

Answer 28

This solution looks so simple, but how to get this... T_T

Answer 29

t^2 derive to 2t

Answer 30

Yea, but then?!

Answer 31

unless youve got an eye for it, lets work a power series

Answer 32

let x = t^2 + c

4t^2

-2t*2t = 0

2 + 4c = 0 is all thats left

c = -1/2 would fit, right?

Answer 33

oh i forgot the initials .... brain slipped a gear

Answer 34

How do you know "x = t^2 + c"??

Answer 35

the wolf, i told you i peeked, and we could most likely guess it if we were brilliant ... but i know im not.

Answer 36

We don't have wolf in exam lol

Answer 37

Let \(\phi = \sum_{k=0}^\infty a_k(t)^k\) be the solution.
\(\phi' = \sum_{k=1}^\infty ka_k(t)^{k-1}\)
\(\phi'' = \sum_{k=2}^\infty k(k-1)a_k(t)^{k-2}\)

x'' - 2tx + 4x
\[=\sum_{k=2}^\infty k(k-1)a_k(t)^{k-2} - 2t\sum_{k=1}^\infty ka_k(t)^{k-1} + 4\sum_{k=0}^\infty a_k(t)^k\]\[=\sum_{k=2}^\infty k(k-1)a_k(t)^{k-2} - 2\sum_{k=1}^\infty ka_k(t)^{k} + 4\sum_{k=0}^\infty a_k(t)^k\]\[=\sum_{k=0}^\infty (k+2)(k+1)a_k(t)^{k} - 2\sum_{k=1}^\infty ka_k(t)^{k} + 4\sum_{k=0}^\infty a_k(t)^k\]\[=2a_2+4a_0 + \sum_{k=0}^\infty [(k+2)(k+1)a_k(t)^{k} - 2ka_k+4a_k ]t^k\]

Answer 38

Oh, the last line should be
\[=2a_2+4a_0 + \sum_{k=1}^\infty [(k+2)(k+1)a_k(t)^{k} - 2ka_k+4a_k ]t^k\]

Answer 39

this is basically a short-cut to solving via series expansions:

repeatedly differentiate:
\[x'' - 2tx' + 4x = 0\\x'''-2tx''+2x'=0\\x^{(4)}-2tx'''=0\\\dots\]
now consider at \(x=0\) we have \(x=-1, x'=0\); it's easy to solve to see \(x''=4\) and then from there \(x'''=0\). any furhter derivatives are then trivially going to be such that \(x^{(n)}=0\) for \(n>4\), so for our analytic solution about \(x=0\) we've determined it is at most a quadratic polynomial \(c_0+c_1 x+c_2 x^2\). it's then a matter of solving \(c_0,c_1,c_2\) using \(x=-1,x'=0,x''=4\) at \(x=0\)

Answer 40

x(0) = -1 => a0 = -1

Answer 41

and x'(0) = 0 => a1 = 0

Answer 42

\[4a_0 + 2a_2 = 0\] => a_2 = 2

Answer 43

Then we have, for the first three terms, -1 + 0(t) + 2(t^2)

Answer 44

So, now, we have the answer from wolf: 2t^2-1, but then, how do we know the rest of the term in the series are all 0 :S

Answer 45

@RolyPoly \[[(k+2)(k+1) - 2k + 4]a_k=0\]the only way to guarantee this expression is \(0\) for all integers \(k\) is for \(a_k=0\)

Answer 46

and the expression msut be \(0\) because it's the coefficient of \(t^k\) in the combined power series expression for \(x''-2tx'+4x\) and our original equation states \(x''-2tx'+4x=0\)

Answer 47

Oh, right!!!
One more question, what do you men when you say "analytic solution", in particular, the word "analytic"?

Answer 48

Also, do we need to show that a is an ordinary point?!

Answer 49

The Laplace Transform of t.f(t).

http://math.stackexchange.com/questions/366802/laplace-transform-of-tft