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In this particular case one would consider the square root of a fractional function of the variable as fairly “ugly”, so why not call this the new variable u? ♦. Applying this substitution we get the integral 1 t(t + 1) 2 = =2 =2 1 − u2 · 1 + u2 t−1 dt = t+1 2 1 1+u2 1−u2 +1 2 4u 1−u 1−u · · 2 du = 1 + u2 2 (1 − u2 ) 1− 1 1 + u2 ·u· 4u (1 − t−1 t+1 du, u= 2 2u du, 1 + u2 du = 2u − 2 Arctan u, t−1 − 2 Arctan t+1 2 u2 ) u= u= t−1 t+1 t−1 t+1 t−1 t+1 . The complete solution is then x = 2t t−1 − 2t Arctan t+1 t−1 t+1 + c t, t > 1, c ∈ R.

T Second solution. Integrating factor. When the equation is multiplied by t > 0 and then read from the right towards the left, we get by a small reformulation that −2t3 = t dx dt d dx +1·x=t· + ·x= (t x), dt dt dt dt where we have used the rule of differentiation of a product. Then by an integration tx = − 1 2t3 dt + c = − t4 + c, 2 and the complete solution is obtained by a division by t > 0: 1 c x = f (t) = − t3 + , 2 t c ∈ R, t > 0. 1 1 When t = 1, we get f (1) = −1 = − + c, so c = − . Therefore, our solution becomes 2 2 x = f (t) = − 1 2 t3 + C.

X= 1 5 t + c t2 , 3 c ∈ R, t < 0. Second solution. A nice little reformulation. When we read the equation from the right towards the left and divide it by t3 = 0, we get by a small rearrangement that 2 d 1 dx 1 dx − 3 ·x= 2 + 2 t dt t t dt dt t2 = 1 t2 ·x= d dt x . t2 Therefore by an integration, x = t2 t2 dt + c = 1 3 t + c, 3 and the complete solution is x= 1 5 t + c · t2 , 3 c ∈ R, t < 0. com 43 Calculus 1c-1 Linear differential equation of first order Third solution. Guess a polynomial. dx and 2x are polynomials of the same degree.

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