A problem in the spectral theory of an ordinary differential by V. A. Tkachenko

By V. A. Tkachenko

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The characteristic polynomial R 2 + 2R + 1 = (R + 1)2 has R = −1 as a double root. The complete solution of the corresponding homogeneous equation is x = c1 e−u + c2 ue−u , u ∈ R; c1 , c2 arbitrary. We can find a particular solution in various ways: a) Guessing. Suppose that x = u2 e−u . Then dx = (−u2 + 2u)e−u du Then by insertion, and d2 x = (u2 − 4u + 2)e−u . du2 d2 x dx + x = (u2 − 4u + 2)e−u + 2(−u2 + 2u)e−u + u2 e−u = 2e−u , +2 du2 du proving that x = u2 e−u is a particular solution. b) Alternative solution.

Hereby we obtain the equivalent equation 2 t = = 1 d2 x 2 2 dx + 3x= − 2 t dt2 t dt t 1 d d 1 dx − 2x = dt t dt t dt 1 d dx 1 dx d 1 d 1 dx + x + − 2 t dt dt dt t dt t dt dt t2 1 dx d 1 d2 x + ·x = 2 , t dt dt t dt t thus d2 x 2 = . 2 dt t t When this is integrated we get d x = 2 ln t + c2 , dt t c1 ∈ R, t ∈ R, hence by another integration, x t = c2 t + c 1 + 2 ln t · 1 dt = c2 t + c1 + 2t · ln t − 2t = 2t · ln t + c1 t + c2 t2 , c1 , c2 ∈ R; t ∈ R? The complete solution is x = 2t2 ln t + c1 t + c2 t2 , c1 , c2 ∈ R; t ∈ R+ .

2 dt t t When this is integrated we get d x = 2 ln t + c2 , dt t c1 ∈ R, t ∈ R, hence by another integration, x t = c2 t + c 1 + 2 ln t · 1 dt = c2 t + c1 + 2t · ln t − 2t = 2t · ln t + c1 t + c2 t2 , c1 , c2 ∈ R; t ∈ R? The complete solution is x = 2t2 ln t + c1 t + c2 t2 , c1 , c2 ∈ R; t ∈ R+ . 5 Consider the differential equation 2t d2 y dy + 2y = 0, + (6 + t) dt2 dt t ∈ R+ . Prove that y = t−2 , t ∈ R+ , is a solution, and then find the complete solution. If we put y = t−2 into the differential equation, we get 2t · (−2) · (−3)t−4 +(6+t) · (−2)t−3 +2t−2 = {12−12}t−3 +{−2+2}t−2 = 0, proving that y1 = t−2 is a solution.

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