View Full Version : Unique solution of 1st order autonomous, homogeneous DE
Pietair
Sep20-10, 10:11 AM
Hello,
1st order autonomous, homogeneous differential equation have the general form:
\dot{x}(t)=ax(t)
It can be shown that the unique solution is always:
x(t)=e^{at}x(t_{0})
I don't get this, could anyone help me?
Thanks!
HallsofIvy
Sep20-10, 11:15 AM
Have you tried at all? What is the derivative of e^{at}. Since x(t_0) is a constant, what is the derivative of e^{at}x(t_0)?
Pietair
Sep20-10, 12:43 PM
Thank you for your answer.
I can work it out when x(t) = x, but this is not the case, is it?
Office_Shredder
Sep20-10, 01:15 PM
x(t) is supposed to be a function of t. What do you mean by x(t)=x?
Pietair
Sep20-10, 01:19 PM
How can I calculate the integral of x(t) when I don't know the corresponding function? x(t) can equal (t^2) or (t-3) and so on, right?
HallsofIvy
Sep21-10, 07:30 AM
Will you please answer my questions? Do you know what the derivative of e^{at} is? Do you know what the derivative of e^{at}x(t_0) is?
HallsofIvy
Sep21-10, 07:32 AM
How can I calculate the integral of x(t) when I don't know the corresponding function? x(t) can equal (t^2) or (t-3) and so on, right?
You can't and you don't want to.
If \frac{dx}{dt}= ax then
\frac{dx}{x}= adt
Integrate both sides of
\int \frac{dx}{x}= \int a dt
Note that on the left you have NO "t". The only variable is x.
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