Author Topic: MT Problem 3  (Read 4027 times)

Victor Ivrii

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MT Problem 3
« on: March 06, 2013, 09:08:26 PM »
Find a particular solution of equation
\begin{equation*}
t^2 y''-2t y' +2y=t^3 e^t.
\end{equation*}

[BONUS] Explain whether the method of undetermined  coefficients to find a particular solution of this equation applies.
« Last Edit: March 06, 2013, 09:13:42 PM by Victor Ivrii »

Jeong Yeon Yook

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Re: MT Problem 3
« Reply #1 on: March 06, 2013, 10:30:47 PM »
The method of undetermined coefficient applies because it only "requires us to make an initial assumption about the form of the particular solution, but with the coefficients left unspecified" (Textbook 10th Edition P.177).

If t = 0, we have, 2y = 0. => y = 0 is the solution for t = 0.
« Last Edit: March 06, 2013, 10:36:17 PM by J. Y. Yook »

Rudolf-Harri Oberg

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Re: MT Problem 3
« Reply #2 on: March 06, 2013, 10:50:34 PM »
This is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version:
$$y''-3y'+2y=0$$

Solving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$.
So, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions.
To use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is:
$c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.
We now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$.

$$c_1'=e^t \implies c_1=e^t$$
$$c_2'=-te^t \implies c_2=-e^t(t-1)$$

Plugging these expressions back to $Y_{gen.hom}$ yields the solution which is
$y=te^t$
« Last Edit: March 06, 2013, 11:33:33 PM by Rudolf-Harri Oberg »

Branden Zipplinger

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Re: MT Problem 3
« Reply #3 on: March 06, 2013, 11:20:47 PM »
for the bonus, the method of undetermined coefficients does not apply here, because when we assume y is of the form g(x), deriving twice and substituting into the equation yields terms with powers of t such that it is impossible to find a coefficient where the solution is of the form you assumed. this can be easily verified

Branden Zipplinger

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Re: MT Problem 3
« Reply #4 on: March 06, 2013, 11:22:05 PM »
(by g(x) i mean the non-homogeneous term)

Brian Bi

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Re: MT Problem 3
« Reply #5 on: March 07, 2013, 12:19:03 AM »
I wrote that undetermined coefficients does not apply because the ODE does not have constant coefficients.

Victor Lam

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Re: MT Problem 3
« Reply #6 on: March 07, 2013, 12:38:24 AM »
I basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this?

Branden Zipplinger

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Re: MT Problem 3
« Reply #7 on: March 07, 2013, 02:20:51 AM »
nevermind.

Victor Ivrii

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Re: MT Problem 3
« Reply #8 on: March 07, 2013, 04:47:03 AM »
Rudolf-Harri Oberg solution is perfect. One does not need to reduce it to constant coefficients (appealing to it is another matter); characteristic equation is $r(r-1)-2r+2=0$ rendering $r_{1,2}=1,2$ and $y_1=t$, $y_2=t^2$ (Euler equation).

Method of undetermined coefficients should not work;  all explanations are almost correct: for equations with constant coefficients the r.h.e. must be of the form $P(x)e^{rx}$ where $P(x)$ is a polynomial but for Euler equation which we have it must be $P(\ln (t)) t^r$ (appeal to reduction) which is not the case.

However sometimes work methods which should not and J. Y. Yook has shown this. Luck sometimes smiles to foolish and ignores the smarts

Quote
Everybody knows that something can't be done and then somebody turns up and he doesn't know it can't be done and he does it.(A. Einstein)

Branden Zipplinger

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Re: MT Problem 3
« Reply #9 on: March 07, 2013, 04:58:04 AM »
has a theorem been discovered that describes what the form of a non-homogeneous equation should look like for it to be solvable by undetermined coefficients?

Patrick Guo

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Re: MT Problem 3
« Reply #10 on: March 16, 2013, 12:51:20 PM »
Just got my midterm back on Friday and looked carefully through..


In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have
   v1 = - ∫ (t^2 + 1) g(t) / Wronskian  dt   ??  what is (t^2 +1) ?! Should that not be y2 = t^2 ?!

And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?

I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.

And why Wronskian = -t^2 ? should it not be t^2 ?

Victor Ivrii

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Re: MT Problem 3
« Reply #11 on: March 16, 2013, 03:00:51 PM »
Just got my midterm back on Friday and looked carefully through..


In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have
   v1 = - ∫ (t^2 + 1) g(t) / Wronskian  dt   ??  what is (t^2 +1) ?! Should that not be y2 = t^2 ?!

And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?

I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.

And why Wronskian = -t^2 ? should it not be t^2 ?

Rats! The answer is simple: typo by the person who typed and the lack of proofreading (all instructors were busy preparing Final and TT2). Thanks! Fixed in all three instances (including on BlackBoard)