Question

differential equations homework help

Answer 1

Find an interval around x = 0 for which the initial value problem: \((x-2)y''+3y=x\), y(0) = 0, y'(0) = 1 has a unique solution.

Answer 2

we can please discuss this. i dont want answer maybe guidaeance

Answer 3

well, in order for this to have a unique solution, i recall that the leading term would have to NOT be zero; so we have a split in the field at x=2

i see no other foibles at the moment

Answer 4

so for guess of *y* can i use \[y=\sum_{n=0}^{\infty} C_n(x-2)^n\]

Answer 5

im not too sure at the moment how the (x-2) would differ in regards to just (x)

Answer 6

but yes, you could try that for a guess

Answer 7

oh. i know u can make 3 guess i think \[C_nx^n\] and \[C_n(x-x_0)^n\] and \[C_nx^{r+n}\] how do u tell when u use number 2 or 3?

Answer 8

my thought is, since x=0 does not create a "bad" setup; that this would be similar to finding the radius or interval of convergence of a power series

Answer 9

oic let me try first can u please check my work if u have time sometime today? i have exam tomorrow so i want to really be good at understand. and thank u for answer me :)

Answer 10

ill be able to check the math, but im not sure if i would be able to check the concepts :)

Answer 11

yes just math hehe series r hard

Answer 12

also one more do i leave the x there? or do change it?

(x-2)y''+3y =-- > x <--

Answer 13

\[(x-2)\sum_2a_nn(n-1)x^{n-2}+\sum_03a_nx^n=x\]
\[(x-2)\sum_2a_nn(n-1)x^{n-2}+\sum_23a_{n-2}x^{n-2}=x\]
\[(x-2)\sum_2[a_nn(n-1)+3a_{n-2}]x^{n-2}=x\]
\[\sum_2[-2a_nn(n-1)-6a_{n-2}]x^{n-2}\\~~~~~~~+\sum_2[a_nn(n-1)+3a_{n-2}]x^{n-1}=x\]
\[\sum_2[-2a_nn(n-1)-6a_{n-2}]x^{n-2}\\~~~~~~~+\sum_3[a_nn(n-2)+3a_{n-3}]x^{n-2}=x\]
\[[-2a_22(2-1)-6a_{2-2}]x^{2-2}\\~~~~~~~+\sum_3[-2a_nn(n-1)-6a_{n-2}]x^{n-2}\\~~~~~~~~~~~~~~~+\sum_3[a_nn(n-2)+3a_{n-3}]x^{n-2}=x\]
\[-4a_2-6a_{0}\\~~~~~~+\sum_3[-2a_nn(n-1)-6a_{n-2}+a_nn(n-2)+3a_{n-3}]x^{n-2}=x\]
\[\sum_3[-2a_nn(n-1)-6a_{n-2}+a_nn(n-2)+3a_{n-3}]x^{n-2}=(4a_2+6a_{0})+x\]

Answer 14

might have gone the wrong way on my index change ....

Answer 15

so i get right now \[y'' = \sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}\] and use \[y= \sum_{n=0}^{\infty}C_nx^n\]
after i get
\[(x-2)\sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}-3\sum_{n=0}^{\infty}C_nx^n\] so i changes it to \[\sum_{n=0}^{\infty}n(n-1)C_nx^{n-1}-2\sum_{n=0}^{\infty}n(n-1)C_nx^{n-2}+3\sum_{n=0}^{\infty}c_nx^n\]

Answer 16

now i will do index change but wat do i change? or how do i do so? because i have n-1 and n-2? do i change all to n only?

Answer 17

bring all down to zero?

Answer 18

to change in index:\[\Large \sum_{n=(3)}x^{(n-k)}=\sum_{n=(3+x)}x^{(n-k-x)}\]

Answer 19

ok so \[\sum_{n=1}^{\infty}n(n+1)c_{n+1}x^n-2\sum_{n=2}^{\infty}(n+2)(n+1)c_{n+2}x^n+3\sum_{n=0}^{\infty}c_nx^n\] now i can factor 1 and 2 terms to bring back to zero?

Answer 20

when we have the x^(n) all lined up, we can factor yes; but we first need to move the indexes to match; so i believe its best to start poping out the ns until n=3 on all of them

Answer 21

well, n=2 in this case

Answer 22

why n=2? if i may ask.

Answer 23

that is where i get confused

Answer 24

\[(x-2)\sum_{2}n(n-1)C_n(x-2)^{n-2}+3\sum_{n=0}C_n(x-2)^n=x\]
\[\sum_{2}n(n-1)C_n(x-2)^{n-1}+3\sum_{n=0}C_n(x-2)^n=x\]
\[\sum_{2}n(n-1)C_n(x-2)^{n-1}+3\sum_{n=0+1}C_{n-1}(x-2)^{n-1}=x\]
\[\sum_{2}[n(n-1)C_n+2C_{n-1}]~(x-2)^{n-1}+3C_{0}=x\]

Answer 25

well, in your setup up there, your highest index is n=2; so we would want to pull out the n=1, and n=2 from the other summations to line everything up

Answer 26

in other words; we know that from n=2 and above, we have a factorable form of the summation; by pulling out the lesser ns to line up our factors, we create seperate terms that can then be able to define a polynomial on the right hand side of it all to compare with

Answer 27

oh did u use \[\sum_{n=0}^{\infty} c_n(x-2)^n\] for u guess?

Answer 28

i gave it a shot, i figured there was no harm yet :)

Answer 29

did it make easier? hehe

Answer 30

as long as the interval of convergence of the power series include x=0, we cshould be fine

Answer 31

\[\sum_{2}[n(n-1)C_n+2C_{n-1}]~(x-2)^{n-1}+3C_{0}=x\]
\[(2C_2+2C_{1})~(x-2)+\sum_{3}[n(n-1)C_n+2C_{n-1}]~(x-2)^{n-1}+3C_{0}=x\]
\[2xC_2+2xC_{1}-4C_2-4C_{1}+\sum_{3}[n(n-1)C_n+2C_{n-1}]~(x-2)^{n-1}+3C_{0}=x\]

Answer 32

notice that we have all we need to define a recurrsion now

Answer 33

\[(3C_0-4C_1-4C_2)+(2C_1+2C_2)x=0+1x\]
and the summation fills with zeros

Answer 34

OHHHHH!!!! I see what you did!! beautiful!!

Answer 35

where does the \[2C_2+2C_1\] come from?

Answer 36

since latex and mathing dont mix that well, you might want to rework it all out by the steps to make sure the end is correct

Answer 37

that comes from pulling out the first few terms of the lower summations as you get the indexes to line up

Answer 38

yes. i am going to try ur method. this worksheet i have is old exam practice question teacher gave no end answers to check.

Answer 39

\[\sum_0a_nx^n+\sum_1b_nx^n+\sum_2c_nx^n\]
\[a_0x^0+b_1x^1+\sum_1a_nx^n+\sum_2b_nx^n+\sum_2c_nx^n\]
\[a_0x^0+b_1x^1+a_1x^1+\sum_2a_nx^n+\sum_2b_nx^n+\sum_2c_nx^n\]
\[a_0x^0+b_1x^1+a_1x^1+\sum_2(a_n+b_n+c_n)x^n\]

Answer 40

in your setup, we end up with a polynomial on the left and right that have to equal, so we compare their coeffs

Answer 41

dont u need to factor 2 for \[a_n\] ? because it begins at zero? i think that is what someone told me

Answer 42

and to match index at 2, u need 2 terms?

Answer 43

correct, tho im not sure how your using the phrase "factor". But yes, you have to remove the first few tems of the series to line up the indexes in order to merge them all into one cohesive summation

Answer 44

\[\large \sum_1^ka_n=a_1+a_2+a_3+...+a_k\]\[a_1+a_2+(a_3+...+a_k)=a_1+a_2+\sum_{3}^{k}a_n\]

Answer 45

so does not matter how many terms? since indez for first begins at zero dont i need 2 terms to pull out?

Answer 46

you want to match your highest index; so the number of terms you pull out will depend on where your other indexes actually start at.

spose your highest index is 4;

an index of 0 will pull out 4 terms
an index of 1 will pull out 3 terms
an index of 2 will pull out 2 terms
an index of 3 will pull out 1 terms

Answer 47

ok. i must go now for a few i will try this problem from start. i still need to figure out to get final solution so i can use y(0) = 0 and y'(0)=0.
but i super appreciate ur help!! if i could i would give you high five and hug! because u really help me well.

Answer 48

if i dont see u later to help me if u can then i will say now have a great day!! :D

Answer 49

so i asked friend about this question. he thinks the answer is much simpler. i dont know if it requires to be solved? just finding interval.

he guesses that interval is \(\infty\) to 2. but why is that?

Answer 50

@amistre64