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366_lab3

# 366_lab3 - (0 1 Є R so there exists a solution ðf/ðx =...

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-2 0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 t x x ' = t x 3 1.

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-2 0 2 4 6 8 10 -4 -3 -2 -1 0 1 2 3 4 t x x ' = 2 x/t 2. a. Parabola, they appear to cross at (0,0). b. tx’ = 2x ∫dx/x=∫2dt/t lnx = 2lnt+c e^(ln(t^2)+c) = x t^2*e^c = x x = ct^2 when t =0, x=0 . It’s not continuous at the point (0, 0).
c. x = ct^2 x(1) = c*1 = c = 8 x = 8t^2 d. x = ct^2 x(0) = c*0 = c = 0 x = 0 e. x = ct^2 x(0) = c*0 = 0 = 8 x does not exist!!!! Because f(t, x) is not continuous at (0, 0), this point is not in the rectangle R. It does not satisfy the hypothesis of existence theorem, so this problem does not have a solution.

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3. f(t, x) = x^2 is defined and continuous on an rectangle R (it is always continuous), and the point

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Unformatted text preview: (0, 1) Є R, so there exists a solution. ðf/ðx = 2x is also always continuous on the rectangle R and the point (0, 1) Є R, so the solution must be unique. t x observation-2 ~ 10 -4 ~ 4 x(2) is outside the window-2 ~ 2 0~15 the curve stops when t is near 1 0.92 ~ 0.95 14.3 ~ 15.1 the curve stops when t = 0.933, cannot find x(2) 4. dx/dt = x^2 ∫dx/x^2 = ∫dt-1/x = t + c x= -1/(t+c) x(0) = -1/c = 1 c = -1 x = -1/(t-1) = 1/(1-t)-2 2 4 6 8 10 5 10 15 20 25 30 t x x ' = x 2 Figure1 Figure2 -2 2 4 6 8 10-4-3-2-1 1 2 3 4 t x x ' = x 2...
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366_lab3 - (0 1 Є R so there exists a solution ðf/ðx =...

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