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)

- anonymous

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- anonymous

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

- amistre64

do you know what a wronskian is?

- anonymous

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

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## More answers

- anonymous

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

- amistre64

or are you at the guessing trial and error stage?

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

what if y2 is a complex function.

- amistre64

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

- amistre64

variation of parameters is what im thinking of

- anonymous

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

- anonymous

or result which is not equal to zero

- amistre64

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

- amistre64

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

- amistre64

any thoughts?

- anonymous

yes...the last method i thought of.

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

I found something.

- amistre64

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

- anonymous

##### 1 Attachment

- anonymous

I think i got it.

##### 1 Attachment

- amistre64

does your solution work back inthe setup?

- 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.

- anonymous

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

- amistre64

it was a heck of a show tho :)

- amistre64

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

- amistre64

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

- anonymous

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

##### 1 Attachment

- amistre64

the wolf ... lol

- amistre64

what is our def of the wronskian?

- anonymous

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

- 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

- 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

- 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 \(y_2(t)=t^3\)

- 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

- 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....

- 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: $$c_1y_1+c_2y_2=0\\c_1y_1'+c_2y_2'=0\\\implies\begin{bmatrix}y_1&y_2\\y_1'&y_2'\end{bmatrix}\begin{bmatrix}c_1\\c_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\\\implies \left|\begin{matrix}y_1(t)&y_2(t)\\y_1'(t)&y_2'(t)\end{matrix}\right|\ne0,\ \forall t\in(0,\infty)\text{ so that no nontrivial solution }c_1,c_2\text{ exists}\\\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{meaning }y_1,y_2\text{ are linearly independent}$$

- 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

- 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 $$n(n-1)-3n+3=0$$

- anonymous

Very good explanation.

- 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.

- anonymous

Thanks @SithsAndGiggles

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