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Change of variable dif equation problem 
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#1
Jul2405, 07:18 PM

P: 12

Driving your 1000 kg car on I5 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.



#2
Jul2405, 07:59 PM

Sci Advisor
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P: 1,123

What have you attempted?



#3
Jul2405, 09:34 PM

P: 12

So far? Nothing...I'm really looking for a place to start =(



#4
Jul2405, 09:44 PM

HW Helper
P: 2,277

Change of variable dif equation problem
Start off with Newton's 2nd Law
[tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex] 


#5
Jul2405, 10:55 PM

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... 


#6
Jul2405, 11:03 PM

HW Helper
P: 2,277

You cannot integrate it like that
[tex] v^3 + 10v = m \frac{dv}{dt} [/tex] [tex] \frac{dt}{m} = \frac{dv}{v^3 + 10v} [/tex] 


#7
Jul2405, 11:25 PM

P: 12

So integrating that I got
t/m = .1 lnv  .05 lnv^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 +/ [140g]^.5 )/2g = v now I'm totally confused 


#8
Jul2405, 11:50 PM

P: 185

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. [tex] \frac{dv}{v^3 + 10v} = m dt [/tex] to integrate it we can breakit into partial fractions [tex] \frac{1}{x^3 + 10x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 10} [/tex] 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. [tex] \int_{v_0}^v \ldots dv = \int_0^t \ldots dt [/tex] 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 [tex] \frac{m}{2} \frac{d}{dt}(v^2) + (v^2)^2 + 10 (v^2) = 0 [/tex] which looks awefully amenable to the substitution u=v^2.. 


#9
Jul2505, 12:12 AM

HW Helper
P: 2,277

You're right Qbert the aceleration should be negative, but you make a slight mistake, It's [itex] \frac{dt}{m} [/itex].



#10
Jul2505, 12:53 AM

P: 12

since m = 1000 (500)u d/dt + u^2 + u = 0 Is it a legal operation to factor a "u" out of this equation? 


#11
Jul2505, 01:33 AM

P: 185

good catch. 


#12
Jul2505, 01:41 AM

P: 185

you have to solve the ODE: m du/dt = (u^2 + 10u). 


#13
Jul2505, 02:27 AM

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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