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)

1. anonymous

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

2. amistre64

do you know what a wronskian is?

3. anonymous

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

4. anonymous

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

5. amistre64

or are you at the guessing trial and error stage?

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. amistre64

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

what if y2 is a complex function.

11. amistre64

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

12. amistre64

variation of parameters is what im thinking of

13. anonymous

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

14. anonymous

or result which is not equal to zero

15. amistre64

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

16. amistre64

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

17. amistre64

any thoughts?

18. anonymous

yes...the last method i thought of.

19. amistre64

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

20. anonymous

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

21. anonymous

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

22. amistre64

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

23. anonymous

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

24. amistre64

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

25. anonymous

I found something.

26. amistre64

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

27. anonymous

28. anonymous

I think i got it.

29. amistre64

does your solution work back inthe setup?

30. anonymous

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

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

32. amistre64

it was a heck of a show tho :)

33. amistre64

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

34. amistre64

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

35. anonymous

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

36. amistre64

the wolf ... lol

37. amistre64

what is our def of the wronskian?

38. anonymous

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

39. amistre64

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

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

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

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

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

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

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

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

Very good explanation.

48. anonymous

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

Thanks @SithsAndGiggles