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18 that e−P (t) = becomes x = = = 1 , and the complete solution 1 + t2 c 1 t2 + 1 dt + 2 2 1+t 1+t 2t2 + 1 c 1 1 1 √ + · 1+ dt 1 + t2 2 1 + t2 1 + ( 2 t)2 √ 1 c 1 1 + · t + √ Arctan( 2t) . com 63 Calculus 1c-1 Linear differential equation of first order Second variant. When we multiply by the integrating factor 1 + t2 we get 1 + t2 d dx 1 + 2t · x = (1 + t2 )x = = (1 + t2 ) 2 1 + 2t dt dt 2 1+ 1 √ 1 + ( 2)2 . Then by an integration, (1 + t2 ) x = 1 2 √ 1 t + √ Arctan( 2t) + c, 2 and the complete solution is given by x= 1 2(1 + t2 ) √ c 1 , t + √ Arctan( 2t) + 1 + t2 2 t ∈ R, c ∈ R.

It is immediately seen that this is true. Alternatively we have p(t) = 1, hence P (t) = t, and thus x = c · e−P (t) = c · e−t . 3, the complete solution is x= 1 (cos t + sin t) + c · e−t , 2 c ∈ R, t ∈ R. Alternatively (sketch only) we get when the equation i multiplied by et that et cos t = et dx d + et x = et x , dt dt and the solution can then be found directly by an integration. 7 Find the complete solution of the diﬀerential equation dx − 2tx = 2t, dt t ∈ R. Show (without applying the Existence and Uniqueness Theorem that there to any (a, b) ∈ R 2 exists precisely one solution, the graph of which goes through (a, b).

E. ( t)3 < 8, from which t < 4. Therefore, 3 3 t ∈ [0, 4[. Finally, since we have assumed that x > 0, x= 1 1 √ , 2 8 − t3/2 t ∈ [0, 4[. 1 Second variant. When we divide by − x3 we get 2 1 −6t 2 = −2x−3 d dx = dt dt 1 x2 . Then by an integration, 1 = −6 x2 1 t 2 dt + 4c = −6 · 3 2 3 t 2 + 4c = −4t 2 + 4c. e. we only get solutions, when c > 0. In that case (x > 0) we have 1 1 1 x= √ = ·√ , 3/2 2 4c − 4t x − t3/2 When (t, x) = c= 1 1, √ 28 t ∈ 0, c2/3 . e. Ic = 0, 82/3 = [0, 4[, and we have found the solution x= 1 1 ·√ , 2 8 − t3/2 t ∈ [0, 4[.