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April 22nd, 2019, 08:55 PM   #1
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Undetermined coefficients

Hello everyone,
In this equation what will be the answer guess for 12t^-2e^(-3t/2)

I know polynomial guess
Trigonometric guess
Exponential guess
But didn’t come across t^negative power
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April 23rd, 2019, 05:21 AM   #2
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This equation can be solved with numerical methods.There may be no elementary solutions.
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April 23rd, 2019, 12:04 PM   #3
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we have a homogeneous solution

$c_1 e^{-3t/2} + c_2 t e^{-3t/2}$

clearly the particular solution won't be exactly one of these forms so let's try

$p(t) = c_3 e^{-3t/2}f_3(t) + c_4 t e^{-3t/2} f_4(t)$

and run it through the diff eq and see what we see.

$4p''(t)+12p'(t)+9p(t) = 4 e^{-\frac{3 t}{2}} \left(c_3 \text{f3}''(t)+c_4 \left(t \text{f4}''(t)+2 \text{f4}'(t)\right)\right)$

It appears we can simplify things by setting $c_4=0$ to obtain

$4c_3e^{-3t/2}f3''(t) = 12t^{-2}e^{-3t/2}$

This gets you a differential equation in $f(t)$ which is easily solved
by repeated integration.

I'm going to let you grind through it all but it appears that a solution that works is

$p(t) =-3 e^{-3t/2}(c t+\ln (t)-3),~c \in \mathbb{R}$
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April 23rd, 2019, 04:26 PM   #4
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$(4y'' + 12y' + 9y)e^{3t/2} = 12t^{-2}$
Integrating and dividing by 4 gives $(y' + (3/2)y)e^{3t/2} = -3/t + \text{c}_1$.
Integrating again gives $ye^{3t/2} = -3\ln|t| + \text{c}_1t + \text{c}_2$.
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April 24th, 2019, 06:55 AM   #5
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I'm going to suggest that, if you can't think of the template to use for the Method of Undetermined Coefficients, you should be using the Method of Variation of Parameters.
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April 24th, 2019, 08:21 AM   #6
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Quote:
Originally Posted by idontknow View Post
This equation can be solved with numerical methods.There may be no elementary solutions.
My mistake on reading the attachment.
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