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Calculus 1c-1, Simple Differential Equations I by Mejlbro L.

By Mejlbro L.

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2) Divide by t2 and rewrite the problem as a problem of integration. I. First solution. 2. e. x = 2t2 ln t − 5t + c · t2 , c ∈ R, t > 0. com 44 Calculus 1c-1 Linear differential equation of first order Second solution. Nice reformulations. By division by t2 = 0 for t > 0 we get 2 5 2 d 1 dx 1 dx + 2 = 2 − 3x= 2 + t t t dt t t dt dt 1 t2 ·x= d dt x . t2 Then by an integration, x = t2 2 dt + t 5 5 dt + c = 2 ln t − + c. t2 t The complete solution is x = 2t2 ln t − 5t + c · t2 , c ∈ R, t > 0. C. Test.

E. one guesses that a solution is x = c · t · ln t. We get by insertion dx 1 − x = c · ln t + c − c · ln t = c = 1. 3 that the complete solution is given by x = t · ln t + c · t, t > 0, c ∈ R. 16 Find the complete solution of the differential equation dx 1 + 1+ dt t x= 1 , t t > 0. A. A linear inhomogeneous differential equation of first order. D. Apply the solution formula. 2 0 2 1 3 Figure 26: The graph of the only bounded solution I. From p(t) = 1 + 1 follows that P (t) = t 1+ 1 t 4 1 1 − , t > 0 for c = −1.

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