Akeem asked in Science & MathematicsMathematics · 6 months ago

# Hello is it possible for someone to help me with this differential equation please?

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• Vaman
Lv 7
6 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,

• 6 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.