Question

xy" -y'+xy' + y = x^2e^(2x)

I tried using Euler-Cauchy equation formula, but that doesn't apply to this, or does it ?

Answer 1

Cauchy does what for us?

Answer 2

\[xy" -y'+xy' + y = x^2e^{2x} \]
\[y" -\frac1xy'+y' +\frac1x y = xe^{2x} \]
\[y" +(1-\frac1x)y' +\frac1x y = xe^{2x} \]

Answer 3

hard to remember much past this :) what would be next? Have you learnted Wronskians yet?

Answer 4

I have learned Wronskians

Answer 5

I tried dividing by x as you did, then replacing y= m or r
making Y'' = r^2 or m^2

Answer 6

Basically, how to find y (c) = x and y (p)= x

Answer 7

\[r^2+(1-\frac1x)r +\frac1x = 0\]
\[r=\frac{-(1-\frac1x)\pm\sqrt{(1-\frac1x)^2-\frac4x}}{2}\]
just wondering

Answer 8

im not sure if my idea is proper for "not constant" coeffs

Answer 9

im thinking of trying a power series solution for the homogenous parts :)

Answer 10

I will try the power series, but hmmm,,,,

Answer 11

\[\frac1xy=\sum_0a_nx^{n-1}\]\[(1-\frac1x)y'=\sum_1a_n*nx^{n-1}-\sum_1a_n*nx^n\]\[y''=\sum_2a_n*n(n-1)x^{n-2}\]

right?

Answer 12

adjust things to the smallest exponent, the feed out till we have like indexes

Answer 13

I'm working on the power series, had to refresh my memory as I've been on variatino of parameters for a while. Thank you ! Keep it coming, if you can.

Answer 14

\[\sum_{1}a_{n-1}x^{n-2}\\ \sum_2a_{n-1}*(n-1)x^{n-2}-\sum_3 a_{n-2}(n-2)x^{n-2}\\\sum_2a_n*n(n-1)x^{n-2}\]

\[a_0~x^{-1}+\sum_{2}a_{n-1}x^{n-2}\]
hmm, i wonder if the -1 exponent casues troubles since a poly only has positive exponents

Answer 15

perhaps we can bring the exponents to positive only,

Answer 16

times it by

Answer 17

\[a_0x^{-1}+a_1+\sum_{3}a_{n-1}x^{n-2}\\ a_1+\sum_3 a_{n-1}(n-1)x^{n-2}-\sum_3 a_{n-2}(n-2)x^{n-2}\\2a_2+\sum_3a_nn(n-1)x^{n-2}\]

\[a_0x^{-1}+2a_1+2a_2\\~~~+\sum_{3}~[a_{n-1}+a_{n-1}(n-1)-a_{n-2}(n-2)+a_nn(n-1)x^{n-2}]~x^{n-2}\]
now this is equal to zero only if all the constants (an) are zero; which makes be think that the x^-1 really doesnt matter. but then again it might :)

\[a_{n-1}+a_{n-1}(n-1)-a_{n-2}(n-2)+a_nn(n-1)x^{n-2}=0~:~n=0,1,2,3,...\]

Answer 18

multiplying everything thru by x doesnt alter the recipe for the ans does it?

Answer 19

\[a_{n-1}+a_{n-1}(n-1)-a_{n-2}(n-2)+a_nn(n-1)=0\]
\[a_n=-\frac{a_{n-1}+a_{n-1}(n-1)-a_{n-2}(n-2)}{n(n-1)}\]
\[a_2=-\frac{a_{2-1}+a_{2-1}(2-1)-a_{2-2}(2-2)}{2(2-1)}\]\[a_2=-\frac{2a_{1}}{2}=-a_1\]

Answer 20

Then again, maybe not.

Answer 21

I'm work at now, gonna retry your solution when I get home.

Answer 22

ok, ill keep at it, since im sure i could use the practice :)

Answer 23

the wolf gives an Ei function which tells me that this doesnt look like it has a pretty solution :)

http://www.wolframalpha.com/input/?i=xy%22+-y%27%2Bxy%27+%2B+y+%3D+x%5E2e%5E%282x%29

Answer 24

for a review if you need it, or just for nostalgia; herb gross does an awesome job at explaning power series solutions

http://archive.org/details/CalculusRevisitedPart3-PowerSeriesSolutions

Answer 25

he also does the other methods so the website is a good resource overall