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January 31st, 2012, 03:51 PM  #1 
Newbie Joined: Jan 2012 Posts: 25 Thanks: 0  Ordinary differential equation y' = 2y + 4
Hey guys, I have to solve this by finding a constant solution and writing y(t) = y_h(t) + y_p(t). So far I have: y' = 2y + 4 y_p = at + b (at + b)'  2(at + b) = 4 a  2at  2b = 4. Now I am stuck. Also, my notes say to set y' = 0, but I am not sure what that means exactly. Any help? 
January 31st, 2012, 05:23 PM  #2 
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: Ordinary differential equation y' = 2y + 4
We are given the ODE: If I were after the general solution, I would first arrange it as: then multiply through by the integrating factor : Integrate with respect to t: However, you are instructed to find the solution to the corresponding homogeneous equation and a particular solution: The associated auxiliary equation is: Thus: Using the method of undetermined coefficients, we assume a particular solution of the form: where A is an arbitrary real constant, and thus: Substituting into the ODE gives: Thus, by the principle of superposition, we have: 
January 31st, 2012, 11:57 PM  #3  
Senior Member Joined: Aug 2011 Posts: 334 Thanks: 8  Re: Ordinary differential equation y' = 2y + 4 Quote:
But this gives ony one solution of the ODE, not all. dy/dx=2y+4 = 2(y+2) dy/(y+2) ) =2 dx ln(y+2)=2x+constant y=C*exp(2x) 2  

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differential, equation, ordinary 
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