how do i get a general solution for y''+(x-1)y=e^x near x=1

Question

amistre64 · Answer

near x=1?  i dont think ive ever heard that before.  any ideas as to what it may be talking about?

anonymous · Answer

in the book it said it would be the (x-$$x_{0}$$)

perl · Answer

use power series

amistre64 · Answer

y'' = e is what this reduces to when x=1 ...

anonymous · Answer

I tried power series but cant find a pattern

amistre64 · Answer

so taylor centered at x=1?

anonymous · Answer

i believe so

amistre64 · Answer

$$y=\sum_0 a_n(x-1)^n$$
like that?

anonymous · Answer

for e^x i made it to be e^(x-1)+1 so i can get $$e \sum_{1}(x-1)^{n}/n!$$

anonymous · Answer

yes

amistre64 · Answer

oh i get it, then the (x-1) tactors in and ...

amistre64 · Answer

notice that we have (x-1)y  in the setup,  if our power series has an 'exponent' part of (x-1)^n then it can factor in easier than if it was just x^n

amistre64 · Answer

$$(x-1)y = \sum_0 a_n(x-1)^{n+1}$$
$$y'' = \sum_2 a_nn(n-1)(x-1)^{n-2}$$

anonymous · Answer

ok i got those two. but when i start solving for a1,.....a4 i see no pattern at all.

amistre64 · Answer

not that this is the case, but the rule for an is sometimes the best we can do if there is no elementary function that it actually represents.  Theres some functions that are simply named becasue they are rather useful but have no good pretty formation.

amistre64 · Answer

http://www.wolframalpha.com/input/?i=y%27%27%2B%28x-1%29y%3De%5Ex well ill be, this IS one of those cases

anonymous · Answer

but if thats the case then how would i be able to produce a general solution

amistre64 · Answer

seeing a pattern, maybe split into 2 cases ... sometimes the best case is to describe it if no real good pattern develops.

amistre64 · Answer

how your course expects you to do it, i wouldnt know ...

anonymous · Answer

wow i just read that and it makes very little sense to me.lol.  hmmm

anonymous · Answer

honestly i don't either

amistre64 · Answer

well, lets work it out and see if we can pattern it if we put our heads together.  oterhwsie it may just be off the wall.

amistre64 · Answer

you ever watch any herbert gross videos?  calculus revisted?  thats where i learned this at

anonymous · Answer

no but i can look it up later. probably should to be honest.

amistre64 · Answer

http://video.mit.edu/watch/lecture-6-power-series-solutions-10680/ to be honest he has a kinda creepy smile to start with but other than that this guy is great

anonymous · Answer

i am watching him and your right he is actually pretty good

amistre64 · Answer

around 21:25 he gets into the solution that has no pretty outcome

amistre64 · Answer

$$(x-1)y = \sum_0 a_n(x-1)^{n+1}$$
$$y'' = \sum_2 a_nn(n-1)(x-1)^{n-2}$$
$$\sum_{0+3} a_{n-3}(x-1)^{n+1-3}$$
$$\sum_{3} a_{n-3}(x-1)^{n-2}$$
$$2a_2+\sum_3 a_nn(n-1)(x-1)^{n-2}$$
$$2a_2+\sum_3 [a_nn(n-1)+a_{n-3}](x-1)^{n-2}=e^x $$
you think we shold turn e^x into a power series? and compare term by term?  eventually that is

anonymous · Answer

thats what i tried and got a power series with no visible pattern

amistre64 · Answer

28:50 he starts talking about what happens when we have no pretty outcome and we simply name it if its important enough.

amistre64 · Answer

im wondering, just a thought, if we would solve the homogenous first, and then attempt to get the particular . not sure if that really matters tho

anonymous · Answer

i thought of that but got lost when doing it. well i got to go to class now maybe he will discuss it later. thanks for the help.

amistre64 · Answer

good luck, i believe i heard him say that we can work the homogenous and then variation for the particular.

amistre64 · Answer

either way, the wolf gives some rather bizarre 'solution'

anonymous · Answer

yeah it does . bye and thanks again.

amistre64 · Answer

like usual, solve the homogenous to get a set of functions to work with the particular
$$2a_2+\sum_3 [a_nn(n-1)+a_{n-3}](x-1)^{n-2}=0$$
$$a_0=c_0~:~a_1=c_1~:~a_2=0$$
$$a_n=-a_{n-3}\frac{1}{n(n-1)}~:~n\ge3$$
$$a_3=-a_0\frac{1}{3!}$$
$$a_4=-a_1\frac{2}{4!}$$
$$a_6=a_0\frac{1(4)}{6!}$$
$$a_7=a_1\frac{2(5)}{7!}$$
---------------------------------------
$$\sum_{n=1}^{\infty}a_{3n}(x-1)^{3n}=\sum_{n=1}^{\infty}a_0\frac{(-1)^n\prod_{k=1}^{n}(3k-2)}{(3n)!}(x-1)^{3n}=c_0g(x-1)$$
$$\sum_{n=1}^{\infty}a_{3n+1}(x-1)^{3n+1}=\sum_{n=1}^{\infty}a_1\frac{(-1)^n\prod_{k=1}^{n}(3k-1)}{(3n+1)!}(x-1)^{3n+1}=c_1~h(x-1)$$
----------------------------------------

$$(x-1)y_p=(x-1)A(x-1)g(x-1)+(x-1)B(x-1)h(x-1)$$

$$y''_p=A''(x-1)g(x-1)+A'(x-1)g'(x-1)
+A'(x-1)g'(x-1)+A(x-1)g''(x-1)
+B''(x-1)h(x-1)+B'(x-1)h'(x-1)
+B'(x-1)h'(x-1)+B(x-1)h''(x-1)$$
ill have to dig thru this to see which parts conform to the homogenous and equate to 0 but this is what ive got to so far

amistre64 · Answer

I had to relearn variation of parameters; simply forgot to 'adjust' the process along the way.

$$y_p=Ag+Bh$$
$$y'_p=Ag'+Bh'+(\underbrace{A'g+B'h}_{=0})$$
$$y''_p=A'g'+B'h'+Ag''+Bh''+[(\underbrace{A'g+B'h}_{=0})]'$$
plugging this into the equation get us:
$$\begin{pmatrix}
+Axg&+Bxh&+A'g'\

+Ag''&+Bh''&+B'h'\
\color{red}{=0}&\color{red}{=0}&\
\end{pmatrix}=e^x$$
the first 2 columns have a factor of the homogenous solution
$$A(g''+xg:=0)~:and:~B(h''+xh:=0)$$

so we are left with the system of equations:

$$A'g+B'h=0\
A'g'+B'h'=e^x $$

solving for A' and B' allows us to integrate them back into A and B.

If theres another route to this its beyond my learning so far.  I heard that there is a variation of parameters that doesnt rely on y_h but I havent had the time to review it, maybe tonight ....