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MT Problem 3

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Victor Ivrii:
Find a particular solution of equation
t^2 y''-2t y' +2y=t^3 e^t.

[BONUS] Explain whether the method of undetermined  coefficients to find a particular solution of this equation applies.

Jeong Yeon Yook:
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.

Rudolf-Harri Oberg:
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:

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$c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.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

Branden Zipplinger:
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:
(by g(x) i mean the non-homogeneous term)


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