October 30th, 2011, 01:08 PM  #1 
Newbie Joined: Oct 2011 Posts: 5 Thanks: 0  Variation of parameters?
Find the general solution of the differential equation: y''  3y' = 5sin(3x) roots are 0 and 3, so how do i get a Wronskian from that? Can't use undetermined coeffs because there's no y term.. 
October 30th, 2011, 03:16 PM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Variation of parameters?
Don't just memorize formulas (like the "Jacobian"), learn concepts. Here, the fact that 0 and 3 satisfy the characteristic equation means that solutions to the associated homogeneous equation are and . From that you can find that the Jacobian is . But more important is the idea that we can find a solution to the entire equation of the form for some functions u(x) and v(x). In fact it is easy to see there are an infinite number of such solutionsgiven a solution, y, pick u(x) to be any differentiable function you please, and you can solve for v(x). If , then . Since there exist an infinite number of such solutions, we "narrow the search" by requiring that . That leaves so that . Putting those forms for y' and y'' into the differential equation, we have . Solve that for u', then integrate to find u. Once you know u', you can put that into the previous condition, to solve for v' and integrate to find v. However, I don't know why you say "Can't use undetermined coeffs because there's no y term". Whether there is a y term or not is irrelevant. Try y(x)= Asin(3x)+ Bcos(3x). 
October 31st, 2011, 08:01 AM  #3 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Variation of parameters? [color=#000000]Another way to do it. Particular Solution: Substitute Homogeneous Solution: so .[/color] 

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