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May 13th, 2013, 05:27 AM  #1 
Member Joined: Apr 2013 Posts: 42 Thanks: 1  differential equation
Given is the differential equation where b is constant and real. How do I show that for any the solution to the equation is equal to , where and A and B are constant and real? And I also wonder how to determine for b=2, y(0)=2 and y'(0)=8 the solution to the differential equation? 
May 13th, 2013, 05:36 AM  #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: differential equation
I would multiply the ODE by , then write the characteristic equation and use the quadratic formula to determine the roots. What do you find?

May 13th, 2013, 06:44 AM  #3 
Member Joined: Apr 2013 Posts: 42 Thanks: 1  Re: differential equation
I find . So ? Almost there, only I have a instead of in the sinus and cosinus...

May 13th, 2013, 07:26 AM  #4 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: differential equation
That is a second order, linear homogeneous differential equation. The theory says that the set of all solutions is a vector space of dimension 2 which means that every solution can be written as a linear combination of two functions that satisfy the equation. Clearly is a linear combination of and . So it is sufficient to show that and satisfy the differential equation and are independent.

May 13th, 2013, 08:47 AM  #5 
Member Joined: Apr 2013 Posts: 42 Thanks: 1  Re: differential equation
I forgot to take the square root of . When I do that, the answer is correct. And the second question is just substituting b by 2 and equate y and y' to the given values? It gives me A=2 and B=5/4 

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