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April 12th, 2007, 10:20 PM  #1 
Newbie Joined: Apr 2007 Posts: 6 Thanks: 0  Differential equation
somebody that can help me solve this? find the general solution dy/dt=2y I know that the preposition dy/dt=ay gives y=ce^at and the answer is supposed to be y=ac^2t, but proving this was worse... 
April 13th, 2007, 05:09 AM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
In other words, you are saying that: y' + 2y = 0 If we had initial values, we could solve this by Laplace transformations, but since we don't, we'll have to go by a different route. We'll have to say that r^2 + 2 = 0 and then solve for the complex number r, and then use the two real parts of e^r as the terms that will give us the solution to the problem. 
April 14th, 2007, 01:43 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,484 Thanks: 2041 
Multiply by e^(2t), then integrate.

May 6th, 2007, 03:29 PM  #4 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
dy/dt = 2y multiply with dy and divide with y and you get: dy/y=2dt integrate and: ln(y)ln(y0)=2t ln(y)=2*t+ln(y0) y=exp(2t+ln(y0)) y=y0*exp(2t) where y0 is the initial value! 
May 6th, 2007, 11:36 PM  #5  
Global Moderator Joined: Dec 2006 Posts: 20,484 Thanks: 2041  Quote:
Alternatively, one can solve the characteristic equation, r + 2 = 0 (not r^2 + 2 = 0), and hence write down the answer.  

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