0
$\begingroup$

MIT Open CourseWare 18.03 Spring 10 Exercises 1B-2 c)

My Question: What is the logic and method by which the correct integrating factor was found? I found an exponential function that is not the correct integrating factor.

Find and integrating factor and solve:

$ (t^2 + 4) dt + t dx = x dt $

I put this ODE in form:

$ x -t(\frac{dx}{dt}) = t^2 + 4 $

and found an integrating factor $ e^{-2t^2} $

The solution manual has an integrating factor $ \frac1{t^2} $

What is the process by which this integrating factor was found? I do not understand the algebra. The solutions are at this URL:

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/readings/notes_exe/MIT18_03S10_1ex.pdf

$\endgroup$
6
$\begingroup$

You want a coefficient of $1$ on the derivative. So your ODE should instead look like $\frac{dx}{dt}-\frac{1}{t}x=-t-4/t$. Then your integrating factor would be $e^{-\ln(t)}=1/t$; that converts the equation into $\frac{d}{dt}(x/t)=-1-4/t^2$.

I don't really understand the integrating factor of $1/t^2$ here.

$\endgroup$
  • $\begingroup$ I see my error thanks $\endgroup$ – Anthony Orona Jun 26 '16 at 1:50
0
$\begingroup$

As pointed out by @Ian, the coefficient of the derivative should be 1. Moreover, integrating factor is $ e^{\int (coefficient\space of \space x) dt} $ which is $\frac {1}{t}$ in this case.$\frac{1}{t^2} $ must be a typo.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.