A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Differential Equation: If one fundamental solution of (t^2)y''-(3t)y'+3y=0 is y_1(t) = t, what is another fundamental solution y_2(t) satisfying y_2(1)=1 and y'_2(1)=3. (Using the definition of the Wronskian)

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[t^2y'' - 3ty' + 3y = 0 \] \[y _{1}(t) = t, y _{2}(1)=1, y'_{2}(1)=3. \] Find\[y _{2}(t)\]

  2. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do you know what a wronskian is?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wronksian, w(y_1,y_2) = y1y2' - y2y1'

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was thinking that y = C1t + C2e^(RT)

  5. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    or are you at the guessing trial and error stage?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm trying to figure out the approach. So yes, im at the trial and error

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i tried to do t^2R^2 -3tR +3 = 0, and find R in terms of T.

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    using -b+- sqrt(b^2-4ac) / 2a gives me sqrt(-3t^2) so that did not work as planned.

  9. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let y2 = At, were A is a function of t comes to mind ... but i cant really recall the trial and error stuff

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what if y2 is a complex function.

  11. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    y2 = At y2' = A + A't , let A't = 0 y2'' = A' want sure if you were working with complexes

  12. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    variation of parameters is what im thinking of

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was working it in terms of ty2' - y2 and i know this have to be equal of a specific function..

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or result which is not equal to zero

  15. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    t^2 A' - 3At -3A't + 3At ------------ 0 = t^2 A' if A' = 0, so let A be a constant maybe?

  16. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    y2 = k y1 = t y = y1+y2 y = ct + k

  17. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    any thoughts?

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes...the last method i thought of.

  19. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i cant say im familiar with you last method so im not able to comment on it

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so y1 = t and y2 = e^(RT) ?

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm looking up second order homogeneous diff eQ. Actually, we did not cover variation of parameters.

  22. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    t^2 r^2 e^(rt) - 3t r e^(rt) + 3 e^(rt) ------------- 0 = e^(rt) (t^2 r^2 -3tr +3)

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is my approach here....along with Wronskian... http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

  24. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    tr = (3 +- sqrt(-3))/2 r = (3 +- sqrt(-3))/2t

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I found something.

  26. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    my approach fails the conditions for y2(1) and y2'(1)

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think i got it.

  29. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    does your solution work back inthe setup?

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I used Abel's theorem and based on the equation, the W(y1,y2) = C3/t&+^3 , C3 was used a variable since integrating y2 again will bring forth another variable which is C2 and since I was given y2 and y2' initial condition i used it to substitute and find c2 and c3. I have to try and work backwards.

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it does not :( the w(y1,y2) = 6/t^6 - 4/t it should have been 2/t^3

  32. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it was a heck of a show tho :)

  33. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    y2 = t^3 .... try to work towards that

  34. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    3t^3 -3t(3t^2) t^2 6t 3t^3 -9t^3 +6t^3 = 0

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I attached the problem just as it is. By the way, how did you deduce y2 to be t^3?

  36. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the wolf ... lol

  37. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what is our def of the wronskian?

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1443482195702:dw| Must not equal zero for it to have fundamental solution or be linearly independent.

  39. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    t^2 n(n-1) t^{n-2} -3t n t^{n-1} +3 t^n = 0 n(n-1) t^{n} -3n t^{n} +3 t^{n} = 0 n(n-1) -3n +3 = 0 n^2 -4n +3 = 0 (n-1)(n-3) = 0

  40. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ty' - y not= 0 not sure how using the definition helps us ... as long as y'/y not equal 1/t we are good i spose

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we have $$t^2y''-3ty'+3y=0\\\implies y''-\frac3ty'+\frac3{t^2}=0$$ Abel's theorem tells us that for fundamental solutions \(y_1,y_2\) to our linear differential equation the Wronskian is a constant multiple of \(e^{-\int -3/t\, dt}=e^{3\ln t}=t^3\), so we have $$\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|=C_1t^3\\y_1y_2'-y_1'y_2=C_1t^3$$we're told that \(y_1=t\) and \(y_1'=1\) so $$ty_2'-y_2=C_1t^3\\y_2'-\frac1ty_2=C_1t^2$$ so let's solve this first-order ODE using an integrating factor, \(\mu=e^{-\int\frac1t\, dt}=e^{-\ln t}=\frac1t\) giving us $$\frac1ty_2'-\frac1{t^2}y_2=C_1t\\\left(\frac1ty_2\right)'=C_1t\\\implies \frac1t y_2=\frac{C_1}2t^2+C_2\\\implies y_2(t)=\frac{C_1}2t^3+C_2t$$ now consider that we're told \(y_2(1)=1\) and \(y_2'(1)=3\) so: $$y_2'(t)=\frac32C_1t^2+C_2\\\\y_2(1)=1\implies \frac12C_1+C_2=1\\y_2'(1)=3\implies\frac32C_1+C_2=3\\\implies C_1=2,C_2=0$$giving us \(y_2(t)=t^3\)

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now double check the Wronskian: $$\left|\begin{matrix}t&t^3\\1&3t^2\end{matrix}\right|=3t^3-t^3=2t^3$$ and \(C_1=2\) indeed

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I see where I made my mistake. The photo i attached was the correct approach. I took the positive integral instead of the negative as such it became C/t^3 and not Ct^3... Thanks a lot @oldrin.bataku And this kids, is why people hate math....

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    be careful! and yeah @amistre64 if the Wronskian were zero (i.e. \(C_1=0\)) then they would be linearly *dependent*, whereas we want them to be linearly *independent* i.e. \(C_1\ne 0\), so that there is no nontrivial solution to the sytsem of linear equations you get the Wronskian from: $$c_1y_1+c_2y_2=0\\c_1y_1'+c_2y_2'=0\\\implies\begin{bmatrix}y_1&y_2\\y_1'&y_2'\end{bmatrix}\begin{bmatrix}c_1\\c_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\\\implies \left|\begin{matrix}y_1(t)&y_2(t)\\y_1'(t)&y_2'(t)\end{matrix}\right|\ne0,\ \forall t\in(0,\infty)\text{ so that no nontrivial solution }c_1,c_2\text{ exists}\\\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{meaning }y_1,y_2\text{ are linearly independent}$$

  45. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and we're looking at \((0,\infty)\) because we were given conditions on \(t=1\) and our differential equation has a singular point at \(t=0\) so the solution on \((-\infty,0)\) is undetermined

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    also @amistre64 found another fundamental solution using the fact that it's a Cauchy-Euler equation, where we can use a clever change of variables \(x=\ln t\) to reduce it to a constant-coefficient problem and the characteristic equation is $$n(n-1)-3n+3=0$$

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Very good explanation.

  48. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's also possible you were expected to solve via reduction of order. using the known solution \(y_1=t\), you suppose \(y_2=y_1v=tv\) is a solution. You have \[{y_2}'=v+tv'\quad\text{and}\quad{y_2}''=2v'+tv''\] Substituting into the ODE, you have \[t^2y''-3ty'+3y=t^2(2v'+tv'')-3t(v+tv')+3tv=0\] Simplifying some, you get \[t^3v''+(2t^2-t)v'=0\] Substitute \(x=v'\): \[t^3x'+(2t^2-t)x=t^2x'+(2t-1)x=0\] which is linear and separable.

  49. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks @SithsAndGiggles

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.