Solving an Equation using Power SeriesDate: 05/18/99 at 10:25:03 From: Heinz Subject: Diff. Equation Hello Dr Math, I'm stuck with this problem: Solve the differential equation (t^4)*x'' + x = 0 for x. How could one go about solving an equation of this sort? I would greatly appreciate your help. Sincerely, Heinz Date: 05/18/99 at 12:58:08 From: Doctor Nick Subject: Re: Diff. Equation Hi Heinz - The way to solve an equation like this is with power series. Assume the solution has a power series in t around some point (for instance, 0). Differentiating the power series twice, multiplying by t^4, and using the differential equation will give you a recurrence relation for the coefficients in the power series; that is, you'll get an expression that allows you to give the n-th coefficient as a function of earlier coefficients. Take a look at just about any book on differential equations. There should be a chapter on power series solutions. It's quite a powerful method. Feel free to write back if you need more help. - Doctor Nick, The Math Forum http://mathforum.org/dr.math/ Date: 05/18/99 at 13:27:50 From: Doctor Rob Subject: Re: Diff. Equation Thanks for writing to Ask Dr. Math! A good approach is to regard x as a function of v = 1/t. Then x' = dx/dt = (dx/dv)*(dv/dt) = (-1/t^2)*(dx/dv) x" = dx'/dt = (1/t^4)*(d^2x/dv^2) + (2/t^3)*(dx/dv) so the equation for x in terms of v becomes d^2x/dv^2 + (2/v)*(dx/dv) + x = 0 Now write x = y/v, which will have the effect of clearing fractions, so dx/dv = (v*dy/dv - y)/v^2 = 1/v*dy/dv - y/v^2 d^x*dv^2 = (1/v)*(dy^2/dv^2) - (2/v^2)*(dy/dv) + 2*y/v^3 and the equation is (1/v)*(d^2y/dv^2) + y/v = 0 d^2y/dv^2 + y = 0 (v is not zero because it equals 1/t, and t is not infinite.) Now this is a familiar equation. The solutions have the form y = A*sin(v) + B*cos(v) where A and B are arbitrary constants. Then x = y/v = [A*sin(v) + B*cos(v)]/v x = t*[A*sin(1/t) + B*cos(1/t)] which is the general solution. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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