MAT244-2013S > MidTerm

MT Problem 3

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Brian Bi:
I wrote that undetermined coefficients does not apply because the ODE does not have constant coefficients.

Victor Lam:
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:
nevermind.

Victor Ivrii:
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)

--- End quote ---

Branden Zipplinger:
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?

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