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Mathematics
Calculus
Non solvable integral? (dx/dt)^2 dt
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[QUOTE="Tomder, post: 6865916, member: 734344"] [B]TL;DR Summary:[/B] I have a non linear system to which I implement a PD controller, but when applying the kinetic-work theorem I can't solve an integral. The integral is (dx/dt)^2 dt, where x=x(t) so it can't be just x + C. The non linear system for whom wants to know how did I get to that point is: d(dx/dt)/dt = sqrt(a^2+b^2)*sin(x+alfa+phi) - Kd*(dx/dt); where alfa = atan(a/b), phi = constant angle, Kd = constant coefficient. After applying the kinetic work theorem by multiplying both sides by dx/dt I get: d(dx/dt)/dt *dx/dt = sqrt(a^2+b^2)*sin(x+alfa+phi)*dx/dt - Kd*(dx/dt)*dx/dt ; So, by integration by dt I get to: 1/2*(dx/dt)^2 = - sqrt(a^2+b^2)*cos(x+alfa`phi) - INTEGRAL[(Kd*(dx/dt)^2]dt + C ; Rearranging terms: 1/2*(dx/dt)^2 + sqrt(a^2+b^2)*cos(x+alfa`phi) = C - INTEGRAL[(Kd*(dx/dt)^2]dt ;And by this without the Kd term i could get the total energy of the system and the velocity at every point but I don't know how to proceed with the Kd term. I'm sure maybe some of the theory may be wrong explained so I say sorry in advance. For further explanation, my system doesn't lose energy when Kd = 0 because the total energy would be constant but with Kd term I assume it is like a frictional component that takes the energy out and the system would slowly stop oscillating in the equilibrium point. [/QUOTE]
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Mathematics
Calculus
Non solvable integral? (dx/dt)^2 dt
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