Real functions in one variable. Simple DEs 1 by Mejlbro L.

By Mejlbro L.

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2 7 28 Finally, we get by a differentiation dx 1 1 3√ 1 = · − · − t · √ 3 dt 2 2 2 8 − t3/2 3 √ √ 1 1 √ t = 3x3 t. 2 8 − t3/2 We have checked our solution. 3) According to (1) the general solution has the form = 3 x= 1 1 √ , 2 c − t3/2 t ∈ 0, c2/3 , c > 0. √ When t → c2/3 −, we get c − t3/2 → 0+, from which we conclude that x = f (t) → +∞ for t → c2/3 −, and the solution has the vertical asymptote t = c2/3 . 15 Find the solution x = ϕ(t) of the differential equation dx 2t = x, t ∈ R, x ∈ R, dt e √ for which ϕ( 2) = 0.

The function x = x = f (t) = 1 1 √ , 2 8 − t3/2 t ∈ [0, 4[. Then f (t) is defined in the whole of the interval [0+, 4[. 1 1 1 For t = 1 we get f (1) = √ = √ . 2 7 28 Finally, we get by a differentiation dx 1 1 3√ 1 = · − · − t · √ 3 dt 2 2 2 8 − t3/2 3 √ √ 1 1 √ t = 3x3 t. 2 8 − t3/2 We have checked our solution. 3) According to (1) the general solution has the form = 3 x= 1 1 √ , 2 c − t3/2 t ∈ 0, c2/3 , c > 0. √ When t → c2/3 −, we get c − t3/2 → 0+, from which we conclude that x = f (t) → +∞ for t → c2/3 −, and the solution has the vertical asymptote t = c2/3 .

Com 48 Calculus 1c-1 Linear differential equation of first order Second solution. When the equation is multiplied by the integrating factor exp −t2 , we get by the rule of differentiation of a product that 2 dx − 2t e−t x dt 2 2 d d −t2 dx + e−t · x = x e−t . = e dt dt dt Then by an integration, 2 2t e−t 2 x e−t = 2 = e−t 2 2 2t e−t dt + c = −e−t + c, 2 hence by a multiplication by et = 0, 2 x = −1 + c · et , c ∈ R, t ∈ R. Third solution. Separation of the variables. It follows from dx = 2tx + 2t = 2t(x + 1) dt that the variables can be separated.

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