Differential equation help

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?

Reply 1

econhelp525

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

16376863911598487997127844261910.jpg527.5KB

Reply 2

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.

(edited 2 years ago)

Reply 3

econhelp525

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.

Reply 4

mqb2766

I think the sub error in the working is when you replace x by z. z = x^(-1), not x.

Reply 5

mqb2766

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.

Reply 6

econhelp525

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...

Reply 7

econhelp525

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?

Reply 8

mqb2766

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

(edited 2 years ago)

Reply 9

econhelp525

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!

Reply 10

mqb2766

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.

Reply 11

econhelp525

Got it, I think the answer is:

16376894646083787314181485099446.jpg516.3KB

Reply 12

mqb2766

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.

Reply 13

DFranklin

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...

Reply 14

econhelp525

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.

Reply 15

mqb2766

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.