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anonymous
 one year ago
1st ODE !
anonymous
 one year ago
1st ODE !

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This thing isn't liniar any ideas ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm I have an idea.. but it's not giving me the same solution as Wolfram.. So I'm thinking I made a boo boo somewhere. I'll at least show you my attempt

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let \(\large\rm u=xt\) Differentiating our sub with respect to time gives \(\large\rm u'=x'1\qquad\to\qquad x'=u'+1\) Subbing everything in gives us,\[\large\rm u'+1=(u)^2+1\]\[\large\rm u'=u^2\]And then just seperation, ya? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ooo goodie! I actually am getting the same as wolfram, i just didn't simplify it far enough :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Able to make sense of that? It should be correct :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436763190196:dw Am I right @zepdrix ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1A little hard to read that last line. Lemme just make a note that you should be careful to include your constant of integration at this step.\[\large\rm \frac{1}{u}=t+c\]Put the negative on the other side,\[\large\rm \frac{1}{u}=ct\]Solving for u,\[\large\rm \frac{1}{ct}=u\]Then you need to undo your substitution. Remember, we're trying to solve for x(t).
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