Variation of ParametersDate: 11/10/2000 at 12:06:02 From: Megan Hall Subject: Diff. Eq. and Variation of Parameters I know that y" + 4y = sec(x) is a variation of parameters problem, but I can't seem to solve it. I'm not using the Cramer's Rule form, but the other standard form that involves solving for u and v as a system of equations. Help! Please show me how to solve this one. I think I'm getting lost in solving for u and v and integrating. Date: 11/11/2000 at 11:08:49 From: Doctor Anthony Subject: Re: Diff. Eq. and Variation of Parameters The equation is (D^2 + 4)y = sec(x) The complementary function (CF) is found from the auxiliary equation m^2 + 4 = 0 m = +-2i The CF gives y = a*cos(2x) + b*sin(2x) where a and b are arbitrary constants. For the general solution we use the method of variation of parameters. Since the two linearly independent solutions of the CF are known, we can find a particular solution as follows: y = u1*cos(2x) + u2*sin(2x) where u1 and u2 are unknown functions of x which are to be determined. Then differentiating y' = 2u1(-sin(2x)) + u1'.cos(2x) + 2u2(cos(2x)) + u2'.sin(2x) = -2u1.sin(2x) + 2u2.cos(2x) + u1'.cos(2x) + u2'.sin(2x) y" = etc. Having then substituted into the original differential equation we end up with u1'*cos(2x) + u2'*sin(2x) + 0 = 0 u1'*(-sin(2x)) + u2'*cos(2x) - sec(x)/2 = 0 Solving for u1' and u2' by determinants we get u1' -u2' 1 ----------------- = ------------------ = ----------------- |sin(2x) 0 | | cos(2x) 0 | | cos(2x) sin(2x)| |cos(2x) -sec(x)/2| |-sin(2x) -sec(x)/2| |-sin(2x) cos(2x)| -sin(2x)sec(x)/2 -sin(x) u1' = ---------------- = ------- = -sin(x) 1 1 +cos(2x)sec(x)/2 u2' = ---------------- = cos(x) - (1/2)sec(x) 1 and integrating u1 = cos(x) u2 = sin(x) - (1/2)ln[sec(x) + tan(x)] P.I. is y = u1*cos(2x) + u2*sin(2x) = cos(x)cos(2x) + sin(2x)*[sin(x)-(1/2)ln[sec(x) + tan(x)]] And so the general solution is y = a*cos(2x) + b*sin(2x) + cos(x)cos(2x) + sin(2x)*[sin(x)-(1/2)ln[sec(x) + tan(x)]] - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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