Solving a Differential Equation by SeriesDate: 06/13/99 at 17:55:06 From: Gene Ko Subject: Solving differential equations by series Solve y' = 2xy by series. I know that this is a separable equation whose solution is y = C*e^(x^2). So if this is solved by series, I should be able to get the same y = C*e^(x^2). But I'm not getting that answer. I'm not sure as to why it doesn't work. This is how I tried to do the problem: "inf" represents "infinity." inf inf y = Sum ((c_n)*(x^n)) y' = Sum (n*(c_n)*(x^(n-1))) n=0 n=1 inf inf Sum (n*(c_n)*(x^(n-1))) = 2x * Sum ((c_n)*(x^n)) n=1 n=0 inf inf Sum ((n+1)*(c_(n+1))*(x^n)) = 2x * Sum ((c_n)*(x^n)) n=0 n=0 (n+1)*(c_(n+1)) 2x*(c_n) c_n = --------------- ----> c_(n+1) = -------- 2x n+1 n=0: c_1 = (2x*(c_0)) / 1 n=1: c_2 = (2x*(c_1)) / 2 = ((2x)^2 * (c_0)) / 2 n=2: c_3 = (2x*(c_2)) / 3 = ((2x)^3 * (c_0)) / 6 n=3: c_4 = (2x*(c_3)) / 4 = ((2x)^4 * (c_0)) / 24 etc. Therefore: (2x)^k * c_0 c_k = ------------ , k >= 1 k! Then: 2x 4x^2 8x^3 16x^4 y = C * (1 + -- + ---- + ---- + ----- + ... ) 1! 2! 3! 4! = C * e^(2x). [FINAL INCORRECT SOLUTION] Any help on where I messed up, and how I can fix it so that I get the correct answer of C * e^(x^2), would be greatly appreciated. Date: 06/13/99 at 18:48:37 From: Doctor Anthony Subject: Re: Solving differential equations by series If we assume a solution of the form y = a0 + a1.x + a2.x^2 + a3.x^3 + ... y' = a1 + 2.a2.x + 3.a3.x^2 + ... Substituting into the differential equation a1 + 2a2.x + 3.a3.x^2 + ... = 2x[a0 + a1.x + a2.x^2 + a3.x^3 + ...] a1 + 2a2.x + 3.a3.x^2 + ... = 2a0.x + 2a1.x^2 + 2a2.x^3 + 2a3.x^4 + ... Equating coefficients on the two sides of the equation: a1 = 0 2.a2 = 2a0, so a2 = a0 3.a3 = 2a1 = 0 4.a4 = 2.a2, so a4 = a2/2 If we continue in this way we find that a1 = a3 = a5 = a7 = ... = 0, and a2 = a0, a4 = a0/2!, a6 = a0/3!, and so on. Therefore y = a0[1 + x^2 + x^4/2! + x^6/3! + ...] = a0.e^(x^2) which agrees with the usual analytical solution. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/