Hello is it possible for someone to help me with this differential equation please?
- VamanLv 76 months ago
m dv/dt =-av^2-k dm/dt. Replace time derivatives with space derivatives
m dv/dx dx/dt =- av^2 - k dm/dx dx/dt. Let dx/dt=v
You have now
m dv/dx = -a v - k dm/dx. Here a= 1/2 rho A cd. Now it is easy to integrate. Try it.
dv/dx +a/m v= -k dm/dx. Put the initial conditions and you will get the answer. Otherwise ask me again.
The maximum velocity is when dv/dt=0,
- az_lenderLv 76 months ago
Making the substitutions to obtain the dv/dt = (18000 - 0.15 v^2)/(3000 - 6t) seems to be high-school algebra, so I leave that part to you.
Using Euler's method in Excel seems pointless, given that the D.E. is separable:
dv/(18000 - 0.15v^2) = dt/(3000 - 6t) =>
(20/3)*dv/(1200000 - v^2) = (1/6)*dt/(500-t).
The expression on the left can be integrated either with a trig substitution or with partial fractions, yielding:
[10/(3*sqrt(1200000)]*ln[(sqrt(1200000) + v)/(sqrt(1200000) - v)] + C = (-1/6)*ln(500 - t) =>
[10/(3*sqrt(1200000)]*ln(1) + C = (-1/6)*ln(500) =>
C = -1.03577 approximately.
Also note that [10/(3*sqrt(1200000)] = 0.0030429 approximately, and if you divide that into the other terms, you have
ln[sqrt(1200000 + v)] - ln[sqrt(1200000) - v] = 340.39 - 54.772*ln(500 - t), or
sqrt(1200000+ v)/sqrt(1200000 - v) = [e^340.39]*(500 - t)^(-54.772).
Now you are in a position to plot the result.
The maximum speed may require some more thought, but it obviously can't exceed 1200000 m/s, considering the square root in the denominator on the left.