Why is the integrating factor for this equation not an exponential function?

Question

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

Answer 1

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.

Answer 2

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.