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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)
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)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[t^2y''  3ty' + 3y = 0 \] \[y _{1}(t) = t, y _{2}(1)=1, y'_{2}(1)=3. \] Find\[y _{2}(t)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you know what a wronskian is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wronksian, w(y_1,y_2) = y1y2'  y2y1'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking that y = C1t + C2e^(RT)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1or are you at the guessing trial and error stage?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to figure out the approach. So yes, im at the trial and error

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried to do t^2R^2 3tR +3 = 0, and find R in terms of T.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0using b+ sqrt(b^24ac) / 2a gives me sqrt(3t^2) so that did not work as planned.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1let y2 = At, were A is a function of t comes to mind ... but i cant really recall the trial and error stuff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what if y2 is a complex function.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y2 = At y2' = A + A't , let A't = 0 y2'' = A' want sure if you were working with complexes

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1variation of parameters is what im thinking of

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was working it in terms of ty2'  y2 and i know this have to be equal of a specific function..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or result which is not equal to zero

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1t^2 A'  3At 3A't + 3At  0 = t^2 A' if A' = 0, so let A be a constant maybe?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y2 = k y1 = t y = y1+y2 y = ct + k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes...the last method i thought of.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i cant say im familiar with you last method so im not able to comment on it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so y1 = t and y2 = e^(RT) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm looking up second order homogeneous diff eQ. Actually, we did not cover variation of parameters.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1t^2 r^2 e^(rt)  3t r e^(rt) + 3 e^(rt)  0 = e^(rt) (t^2 r^2 3tr +3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is my approach here....along with Wronskian... http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1tr = (3 + sqrt(3))/2 r = (3 + sqrt(3))/2t

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1my approach fails the conditions for y2(1) and y2'(1)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1does your solution work back inthe setup?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0it does not :( the w(y1,y2) = 6/t^6  4/t it should have been 2/t^3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1it was a heck of a show tho :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y2 = t^3 .... try to work towards that

amistre64
 one year ago
Best ResponseYou've already chosen the best response.13t^3 3t(3t^2) t^2 6t 3t^3 9t^3 +6t^3 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I attached the problem just as it is. By the way, how did you deduce y2 to be t^3?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is our def of the wronskian?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443482195702:dw Must not equal zero for it to have fundamental solution or be linearly independent.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1t^2 n(n1) t^{n2} 3t n t^{n1} +3 t^n = 0 n(n1) t^{n} 3n t^{n} +3 t^{n} = 0 n(n1) 3n +3 = 0 n^2 4n +3 = 0 (n1)(n3) = 0

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ty'  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
 one year ago
Best ResponseYou've already chosen the best response.0we 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 firstorder 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
 one year ago
Best ResponseYou've already chosen the best response.0now double check the Wronskian: $$\left\begin{matrix}t&t^3\\1&3t^2\end{matrix}\right=3t^3t^3=2t^3$$ and \(C_1=2\) indeed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0be 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
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0also @amistre64 found another fundamental solution using the fact that it's a CauchyEuler equation, where we can use a clever change of variables \(x=\ln t\) to reduce it to a constantcoefficient problem and the characteristic equation is $$n(n1)3n+3=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Very good explanation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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 \[t^3v''+(2t^2t)v'=0\] Substitute \(x=v'\): \[t^3x'+(2t^2t)x=t^2x'+(2t1)x=0\] which is linear and separable.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @SithsAndGiggles
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