Differential equation help

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Thread starter 1 week ago
#1
I've gotten the wrong answer, but I'm not sure where I've gone wrong and how to go about it. Any help?
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Thread starter 1 week ago
#2
(Original post by econhelp525)
I've gotten the wrong answer, but I'm not sure where I've gone wrong and how to go about it. Any help?
Here is the question
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1 week ago
#3
Did you try seperation of variables, so xs on one side, ts on the other and integrate.
Note, it does help to see the orginal question.
Last edited by mqb2766; 1 week ago
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Thread starter 1 week ago
#4
(Original post by mqb2766)
Did you try seperation of variables, so xs on one side, ts on the other and integrate.
Note, it does help to see the orginal question.
Sorry, the question is dx/dt+tx=tx^2 , it was written next to (a).

You can't do separation of variables on this question.
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1 week ago
#5
I think the sub error in the working is when you replace x by z. z = x^(-1), not x.
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1 week ago
#6
(Original post by econhelp525)
Sorry, the question is dx/dt+tx=tx^2 , it was written next to (a).

You can't do separation of variables on this question.
Id be surprised if you couldn't, but there are often a few ways to solve the problem.
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Thread starter 1 week ago
#7
(Original post by mqb2766)
Id be surprised if you couldn't, but there are often a few ways to solve the problem.
It's a Bernoulli equation, it's solved via a transformation of variables and an integrating factor, I've just done something wrong. I'll check out my substitution to see if that's it...
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Thread starter 1 week ago
#8
(Original post by mqb2766)
I think the sub error in the working is when you replace x by z. z = x^(-1), not x.
t/x is the same as tx^-1, so I replaced Z for it, so I don't think this is wrong?
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1 week ago
#9
(Original post by econhelp525)
t/x is the same as tx^-1, so I replaced Z for it, so I don't think this is wrong?
tx^-1 = tz
not t/z
Last edited by mqb2766; 1 week ago
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Thread starter 1 week ago
#10
(Original post by mqb2766)
tx^-1 = tz
not t/z
Oh right! Yes, thank you. I'll amend it, and see what I get. Thanks!
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1 week ago
#11
(Original post by econhelp525)
Oh right! Yes, thank you. I'll amend it, and see what I get. Thanks!
No problem. If you did think about it as seperation of variables, you'd quickly spot that the solution must be some simple function of t^2.
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Thread starter 1 week ago
#12
Got it, I think the answer is:
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1 week ago
#13
(Original post by econhelp525)
Got it, I think the answer is:
Looks about right, but sub it back into the original ODE if you want to verify.
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1 week ago
#14
(Original post by econhelp525)
Sorry, the question is dx/dt+tx=tx^2 , it was written next to (a).

You can't do separation of variables on this question.
dx/dt = t(x^2-x) looks pretty separable to me...
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Thread starter 1 week ago
#15
(Original post by DFranklin)
dx/dt = t(x^2-x) looks pretty separable to me...
I get two pretty different answers if I use separable equations, rather than transformation of variables.
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1 week ago
#16
(Original post by econhelp525)
I get two pretty different answers if I use separable equations, rather than transformation of variables.
You should get the "same" answer in both cases (not counting minor identity transformations). For me, you had a x(x-1) and t. The former would integrate to a log in x (fraction, using partial fractions) and the latter to t^2. Then inverse logs and rearrange to express in terms of x.

Neither way should be too complex to do, but sometimes you can spot simple parts of the answer which can help you validate, even if you dont use that method. If you want to get the seperation of variables sorted, just upload what you tried.
Last edited by mqb2766; 1 week ago
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