# change of variable dif equation problem

by spoon
Tags: equation, variable
 P: 12 Driving your 1000 kg car on I-5 at 100 km/h, you suddenly noticed a large group of animals. You slam on a break and this provides a stopping force equal to v^3 plus the friction force of 10v. Set up and solve a differential equation for velocity. You need to find a suitable change of variable.
 HW Helper Sci Advisor P: 1,123 What have you attempted?
 P: 12 So far? Nothing...I'm really looking for a place to start =(
HW Helper
P: 2,274

## change of variable dif equation problem

Start off with Newton's 2nd Law

$$\sum_{i=1}^{n} F_{i} = m \frac{dv}{dt}$$
 P: 12 Alright, so I starting with Newton's equation...I'm thinking: F = m*V = 1000*V then I'm assuming F = Stopping force which is: F = V^3 +10V = 1000V Integrating both sides... (V^4)/4 + 5V^2 = 500V^2 Then solving for V, I got 13.416 But this totally seems wrong to me...
 HW Helper P: 2,274 You cannot integrate it like that $$v^3 + 10v = m \frac{dv}{dt}$$ $$\frac{dt}{m} = \frac{dv}{v^3 + 10v}$$
 P: 12 So integrating that I got t/m = .1 ln|v| - .05 ln|v^2 + 10| I wasn't really sure how to integrate the right hand side on paper, so I used my calculator Solving for v, using m = 1000: e^(t/50) = v/((v^2) + 10) I used a variable g to = e^(t/50) which led to: gv^2 +10g = v I tried using the quadratic after that: (1 +/- [1-40g]^.5 )/2g = v now I'm totally confused
P: 179
 Quote by cyclovenom $$v^3 + 10v = m \frac{dv}{dt}$$
you'll see you need a minus sign.
dv/dt really aught to be getting smaller
not bigger.

mv' = -(v^3 + 10v)

this is seperable as above.
$$-\frac{dv}{v^3 + 10v} = m dt$$

to integrate it we can breakit into partial fractions
$$\frac{1}{x^3 + 10x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 10}$$
where we pick A, B, C to make the equality hold.

Once you've got the easier to integrate sides.
Integrate using the initial conditions.
$$\int_{v_0}^v \ldots dv = \int_0^t \ldots dt$$

then invert the formula to solve for v.

[or if we want to look for a change of vars..]
mv' + v^3 + 10v = 0.
so
m(vv') + v^4 + 10v^2 = 0

$$\frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0$$

which looks awefully amenable to the substitution u=v^2..
 HW Helper P: 2,274 You're right Qbert the aceleration should be negative, but you make a slight mistake, It's $\frac{dt}{m}$.
P: 12
 Quote by qbert [or if we want to look for a change of vars..] mv' + v^3 + 10v = 0. so m(vv') + v^4 + 10v^2 = 0 $$\frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0$$ which looks awefully amenable to the substitution u=v^2..
Following that...
since m = 1000
(500)u d/dt + u^2 + u = 0

Is it a legal operation to factor a "u" out of this equation?
P: 179
 Quote by Cyclovenom You're right Qbert the aceleration should be negative, but you make a slight mistake, It's $\frac{dt}{m}$
doh! hoisted by my own petard.
good catch.
P: 179
 Quote by spoon Following that... since m = 1000 (500)u d/dt + u^2 + u = 0 Is it a legal operation to factor a "u" out of this equation?
no way! (in this context, d/dt standing alone is meaningless.)

you have to solve the ODE: m du/dt = -(u^2 + 10u).
 P: 12 Okay, so I solving m du/dt = -(u^2 + 10u): t/500 + C1 = 1/10 ln(u+10) - 1/10 ln(u) t/50 + C2 = ln((u+10)/u) where C2 = 10*C1 e^(t/50 + C2) = Ce^(t/50)-1 = 10/u u = 10(Ce^(t/50))^-1 v^2 = 10(Ce^(t/50))^(-1) -----> v = +/-[10(Ce^(t/50))^-1]^(1/2) But v should just be the negative right?

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