x(t) = ae^(bt) + ct is soln to ODE d^2/dt^2 = 16e^(4t) / x

dopey9

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Jul 14, 2006
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I want to find the values of the real constants a, b, and c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far

i differentiated x to get abe^(bt) + c
then i differentialiated that again to get (ab^2)*e^(bt)
then i substituted x and x" into the ODE given to get

(ab^2)*e^(bt) = 16e^(4t) / ae^(bt) +ct

this is where i got stuck..am i on the right path..from this part i need to find a b and c..any help thankz
 
what i got

i managed to work out that a=2 b=2 and c=0...........didi anyone else get that???
 
You're almost there... Picking up from your last equation,

\(\displaystyle \L a^2 b^2 e^{2 b t} + c a b^2 t e^{bt}= 16 e^{4t}\)

The only solutions of this equation are given by setting

\(\displaystyle \L a^2 b^2 = 16\)

\(\displaystyle \L 2b = 4\)

\(\displaystyle \L cab^2 = 0\)

because \(\displaystyle \L e^{Kt}\) and \(\displaystyle \L t e^{Lt}\) are linearly independent functions for any values of constants K and L.

That pretty much lets you find a, b, c. Hope that helps.
 
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