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March 22nd, 2011, 04:33 PM  #11 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  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 = 73e^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 ! 
March 22nd, 2011, 05:41 PM  #12 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: 2nd Order Differential Equation
We begin with: Let's go another route with this one. Multiply through by : Now, notice the leftside 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: 
March 24th, 2011, 07:08 AM  #13 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  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 (x1) + c ) Can you point out my errors, I'm sure there will be some ! Thanks 
March 24th, 2011, 09:51 AM  #14 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: 2nd Order Differential Equation
No errors, looks good! 
March 26th, 2011, 02:01 AM  #15 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  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 = 73te^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 rewritten or am I just looking to much into it? Jake 
March 26th, 2011, 04:58 AM  #16 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  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: 
March 26th, 2011, 07:14 AM  #17 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  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?

March 26th, 2011, 11:44 AM  #18 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: 2nd Order Differential Equation
We have: Using the product rule, we get (after simplification): 
March 26th, 2011, 12:47 PM  #19 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  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 
March 26th, 2011, 12:50 PM  #20 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: 2nd Order Differential Equation
After subbing into the original ODE and simplifying, I got: 

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