differential equation what going on?


by jamesd2008
Tags: differential, equation
jamesd2008
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#1
Mar18-09, 12:11 PM
P: 64
Hi could someone explain why this result occurs when solving this differential equation.

dx/dt= xt

1/xdx/dt=t then intergrating both side we get with respect to dt we get,

Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2

I get really confused how the equation gets re-arranged into the above form after the intergration has occured. Any help please?

Thanks
James
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sutupidmath
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#2
Mar18-09, 02:24 PM
P: 1,635
Quote Quote by jamesd2008 View Post

Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2

I get really confused how the equation gets re-arranged into the above form after the intergration has occured. Any help please?

Thanks
James
You are just exponentiating both sides of the equation. But remember that exponential functions and log's are inverse functions so e^lnx=x
jamesd2008
jamesd2008 is offline
#3
Mar18-09, 02:28 PM
P: 64
Thanks but how come it is c*the e and then raised to the power of 2/t^2. Is that just a rule of exponentials?
Please help doing my head in!!

Thanks
James

tiny-tim
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#4
Mar18-09, 02:56 PM
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differential equation what going on?


Quote Quote by jamesd2008 View Post
Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2
Hi James!

No, x = Cet2/2

C is ec.
EliotHijano
EliotHijano is offline
#5
Mar18-09, 04:22 PM
P: 18
Hey James
dx/dt= xt
dx/x = t·dt
Ln(x) = (t^2)/2 + K
x=e t2 /2 +K = e K · e t2 /2
If we say that e^K = C then;
x= C·e t2 /2

I hope it helps
jamesd2008
jamesd2008 is offline
#6
Mar18-09, 08:39 PM
P: 64
Thanks all for your help got it now!! thanks again james : -)


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