anonymous
  • anonymous
Find two power series solutions of the given differential equation about the ordinary point x=0: y''-xy=0
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
fourier series
anonymous
  • anonymous
.....
anonymous
  • anonymous
\[y=\sum_{n=0}^\infty c_nx^n\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[y'=\sum_{n=1}^\infty c_nnx^{n-1}\]
anonymous
  • anonymous
\[y''=\sum_{n=2}^\infty c_n(n)(n-1)x^{n-2}\]
TuringTest
  • TuringTest
oh I need a review on this bookmark*
anonymous
  • anonymous
i'm pretty sure i have it down so you can come check it after \[y''-xy=\sum_{n=2}^\infty c_nn(n-1)x^{n-2}-x\sum_{n=0}^\infty c_nx^n\]
anonymous
  • anonymous
\[\sum_{n=2}^\infty c_nn(n-1)x^{n-2}-\sum_{n=0}^\infty c_nx^{n+1}\]
anonymous
  • anonymous
\[k=n-2\] \[k+2=n\] \[\sum_{k=0}^\infty c_{k+2}(k+2)(k+2-1)x^k-\sum_{n=0}^\infty c_nx^{n+1}=0\]
anonymous
  • anonymous
\[\sum_{k=0}^\infty c_{k+2}(k+2)(k+1)x^k-\sum_{k=1}^\infty c_{k-1}x^k\] where \[k=n+1\] \[k-1=n\]
anonymous
  • anonymous
pull out one term k=0 of first \[2c_2+\sum_{k=1}^\infty c_{k+2}(k+2)(k+1)x^k-\sum_{k=1}^\infty c_{k-1}x^k\]
anonymous
  • anonymous
\[2c_2+\sum_{k=1}^\infty [c_{k+2}(k+2)(k+1)-c_{k-1}]x^k\]
anonymous
  • anonymous
since identity 0=0 the sum and \[2c_2=0\] \[c_2=0\]
anonymous
  • anonymous
\[c_{k+2}(k+2)(k+1)=c_{k-1}\] \[c_{k+2}=\frac{c_{k-1}}{(k+2)(k+1)}\]
anonymous
  • anonymous
for k=1,2,3,4,5,6......
anonymous
  • anonymous
k=1 \[c_3=\frac{c_0}{3*2}\] k=2 \[c_4=\frac{c_1}{4*3}\] for k =3 \[c_5=\frac{c_2}{constant}\] =\[c_2=0\] k=4 \[c_6=\frac{c_3}{6*5)}=\frac{c_0}{6*5*3*2}\]
anonymous
  • anonymous
k=5 \[c_7=\frac{c_4}{7*6}=\frac{c_1}{7*6*4*3}\]
anonymous
  • anonymous
\[c_8=c_5=0\]
anonymous
  • anonymous
@TuringTest look correct so far?
TuringTest
  • TuringTest
like I said: *bookmark I want you to remind me; it's been a while... maybe @experimentX can verify better than me
experimentX
  • experimentX
jeez ... i forgot all those stuffs. i need to review it myself.
TuringTest
  • TuringTest
hm... @Zarkon care to verify a DE series solution?
anonymous
  • anonymous
Is zarkon the computer on lol
experimentX
  • experimentX
these standard equations power solutions look intimidating ...
anonymous
  • anonymous
\[y=c_0+c_1x+0+\frac{c_0}{3*2}x^3+\frac{c_1}{4*3}x^4+0+\frac{c_0}{2*3*5*6}x^6\] \[+\frac{c_1}{3*4*6*7}x^7+0......\]
anonymous
  • anonymous
\[y_1(x)=1+\frac{1}{2*3}x^3+\frac{1}{2*3*5*6}x^6...\] \[y_2(x)=x+\frac{1}{3*4}x^4+\frac{1}{3*4*6*7}x^7.......\]
experimentX
  • experimentX
what kind of function is this http://www.wolframalpha.com/input/?i=y%27%27+-+xy%3D0
experimentX
  • experimentX
Using Maple I found this solution \[ C0+C1*x+(1/6)*C0*x^3+(1/12)*C1*x^4+\\ (1/180)*C0*x^6+(1/504)*C1*x^7+O(x^8) \]
anonymous
  • anonymous
thats what i have
anonymous
  • anonymous
only i left it not multiplied so i can creat a summation
experimentX
  • experimentX
well, i guess you are right ... i never liked power series solution you know!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.