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§2.3 pp. 152-153 2. Three solutions of a certain second-order linear nonhomogeneous equation are ψ1(t) = 1 + e t2 ψ2(t) = 1 + te t2 ψ2(t) = (t + 1)e t2 + 1. Find the general solution to this equation. Solution We know that the general solution is of the form y(x) = ψ + c1y1 + c2y2 where ψ is any solution to the nonhomogeneous equation (so we could take any of the above solutions to be ψ) and where y1 and y2 are two linearly independent solutions to the homogeneous equation. We know that the difference between any two solutions of the nonhomogeneous equation is a solution to the homogeneous equation, so we may take and y1 = ψ3 − ψ2 = e t2 y2 = ψ3 − ψ1 = te t2 are both solutions to the homogeneous equation. Clearly y1 and y2 are not constant multiples of each other, so they are linearly independent. Taking ψ = ψ1, we get the general solution or more simply y(x) = 1 + e t2 + c1e t2 + c2te t2 y(x) = 1 + c1e t2 + c2te t2 .
§2.4 pp. 156-157 2. Find the general solution of the equation Solution In the homogeneous equation y ′′ − 4y ′ + 4y = te 2t . y ′′ − 4y ′ + 4y = 0 we have b 2 − 4ac = (−4) 2 − 4(1)(4) = 0 so the characteristic equation has the double root r = −b/(2a) = 4/2 = 2. Then the general solution to the homogeneous equation is c1y1 + c2y2 where we have and y1 = e 2t y2 = te 2t . By the variation of parameters method, there is a particular solution of the form ψ = u1y1 + u2y2 Since there formulas for u1 and u2 involve the Wronskian, we compute W [y1, y2] = e 2t (e 2t + 2te 2t ) − 2e 2t (te 2t ) = e 4t Now we apply the formulas for u1 and u2 to get and also Then we have the particular solution −gy2 W dt 2t 2t −(te )te = e4t dt = −t 2 dt u1 = = −t 3 /3 gy1 W dt 2t 2t (te )e = e4t dt = t dt u2 = = t 2 /2 ψ = −t 3 /3 e 2t + t 2 /2 te 2t = 1 6 t3 e 2t so the general solution to the nonhomogeneous equation is y(x) = 1 6 t3 e 2t + c1e 2t + c2te 2t .
Page 1: §2.2 pp. 149-150 Math 4430 HW 5 Se

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