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March 22nd, 2011, 04:33 PM   #11
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Re: 2nd Order Differential Equation

the (x^2) should be in there..

That comes from the y=c1 x^2 e^x +c2 e^-x. I see

Ok great thanks.

On another subject, if a problem d^2 x/dt^2 -4 dx/dt = 7-3e^4t

d^2 x/dt^2 -4 dx/dt =0 , Auxillary equation, if it is of the form m^2 -4m =0, What would it be for m=? , as I realise if the auxillary equation is of the form
m^2 -4 then, m= +or- 2. I should know this !
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March 22nd, 2011, 05:41 PM   #12
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Re: 2nd Order Differential Equation

We begin with:



Let's go another route with this one. Multiply through by :



Now, notice the left-side is the derivative of a product:







Now, multiply through by





Note: the 2nd term in the integrand on the right requires integration by parts.





Since is a constant, we may represent the constant simply as .



Now, had we worked this equation by the method of undetermined coefficients, we would first find:



Next, we would assume a particular solution of the form:

(we need the t as a factor on both terms so that neither term is a form of )





Now, substituting these into the original ODE gives:







Equating coefficients:





Thus, we have:



So, putting it together, we have:





Since the parameters are constants, we may write this in the form:

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March 24th, 2011, 07:08 AM   #13
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Re: 2nd Order Differential Equation



x dy/dx -2y = x^4 e^x
Divide by x^3
x/x^3 dy/dx -2y/x^3 = x^4 e^x /x^3

x^-2 dy/dx -2x^-3 y = x e^x

d/dx y/x^2 = x e^x. Exact form
Integrate wrt x

d(y/x^2) =x e^x dx

y = x^2 (e^x (x-1) + c )

Can you point out my errors, I'm sure there will be some ! Thanks
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March 24th, 2011, 09:51 AM   #14
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Re: 2nd Order Differential Equation

No errors, looks good!
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March 26th, 2011, 02:01 AM   #15
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Re: 2nd Order Differential Equation

Ok so, say I had an equation similar to the 2nd order equation you helped me with before apart from the right side being slightly different.

d^2 x/dt^2 -4 dx/dt = 7-3te^4t Very similar to previous one however there is '3t' in there now.

My question is what form would the complimentry function xh(t) be and what could I try the particular solution xp(t) as. Could the right side be re-written or am I just looking to much into it?

Jake
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March 26th, 2011, 04:58 AM   #16
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Re: 2nd Order Differential Equation

Since the left side is the same, the complimentary function would be the same:



For the particular solution, we could use:



Differentiating, substituting into the ODE, and solving for the undetermined coefficients gives:



Adding this to the complimentary function gives:

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March 26th, 2011, 07:14 AM   #17
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Re: 2nd Order Differential Equation

Awesome , but backing up bit I need to get the 1st and 2nd derivative right I'n my mind, I got x'p(t)= A(2+B)4e^4t + C and x''p(t)=8e^4t(A(B+2t)+2A(B+2t)) is this correct?
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March 26th, 2011, 11:44 AM   #18
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Re: 2nd Order Differential Equation

We have:



Using the product rule, we get (after simplification):



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March 26th, 2011, 12:47 PM   #19
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Re: 2nd Order Differential Equation

Ok I was way out. after subbing into the original equation, what is the end equation after simplfying.... Its a long one. So I can compare coefficients myself
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March 26th, 2011, 12:50 PM   #20
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Re: 2nd Order Differential Equation

After subbing into the original ODE and simplifying, I got:

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