So I'm taking an Intro to ODE's class and we recently learned how to use an integrating factor to solve linear differential equations.  So far this is the only method we have learned for solving these.  Below is my understanding of the steps using a sample problem.

Sample problem:  dx/dt + x/(10 + t) = 1; x(0) = 10, t > 0

Step 1:  Solve for the integrating factor (I.F.).  I understand how to do this, although the why is probably over my head.

I.F. = 10 + t

Step 2:  Multiply both sides of the ODE by the I.F.

(10 + t)[dx/dt + x/(10 + t)] = 1(10 + t)

d[x(10 + t)]/dt + x = 10 + t

Step 3:  Integrate.  This is where I get tripped up conceptually.  What is happening to the left side of the equation?  Where did the "x" term go?  I've been told that d[x(10 + t)]/dt is a perfect derivative based on the product rule, but I don't really understand what that means.

x(10 + t) = (1/2)(10 + t)^2 + C

Step 4:  Solve for x.

x = (10 + t)/2 + C/(10 + t)

Step 5:  Use the initial value to solve for C.

C = 50

Solution:  x(t) = (10 + t)/2 + 50/(10 + t)

As mentioned above, what I need help understanding is what is happening to the equation between steps 2 and 3.  There is a term disappearing from the left side of the equation and I can't figure out why.  I understand the formula for the solution but I don't think that will be enough to get me through the semester so I want to know the process.

Fantastic question. I struggled with this a bit when I took ODE's, too, until I worked through one all the way.

From your post, you understand through step 2, so we'll skip that part. The key to step 3, and where the x went, is in, "I've been told that d[x(10 + t)]/dt is a perfect derivative based on the product rule, but I don't really understand what that means."

Let's work through it. Do you remember the product rule from calculus? When taking the derivative of two things multiplied together, you use the product rule. I always remembered it as first times the derivative of the second, plus the second times the derivative of the first. Let's apply it to your problem to see where the x went.

Just before step 3, we have

d[x(10 +t))]/dx + x = 10 + t

On the left side, you want your integrating factor multiplied by x, or whatever variable your equation is not with respect to (in this case the equation is with respect to t.).

That would mean, pre-integration, we have (x(10+t))'
Post integration, we have x(10 +t), which is what you have in step 3.

Why can we do this?

Let's take the derivative of x(10 + t) using the product rule, with respect to t. First times the derivative of the second, plus the second times the derivative of the first, where x is the first, and 10+t is the second.

x * d(10+t)/dt    +    (10 + t) * d(x)/dt

Simplifying

x * 1 + (10 + t) dx / dt

d[x(10 + t)] /dt + x

This proves that when you integrate the left side of your equation, you'll come up with what's written after step 3 ---> integral of d[x(10+t)]/dt + x = x(10 + t)

Soon you'll become comfortable enough with IF's that when they come along, you'll just write the left side as (x * IF)'
When it's time to integrate, just drop the prime, then solve for x. Easy as that. The product rules just PROVES you can do that as a shortcut.

It's a good thing to know WHY you can take the shortcuts. I had numerous math professors who had ways of figuring out who knew WHY on the test, and who was just applying rules.

Thank you; that's a very useful explanation.  So basically, the left side of the equation in step two is the derivative of x(10+t) (in other words, we are working the equation backwards compared to how we are initially taught to deal with these things).  Is it accurate to say that the result in step two is the result of two partial derivatives?  One with respect to x, and the other with respect to t.  I am also taking Calc 3 this year and just learned about partial derivatives so that might explain some of my confusion.  

I was treating x as some constant because dx/dt was with respect to t.  I was scratching my head why integrating d[x(10+t)/dt + x didn't equal x(10+t) + xt since I thought everything was with respect to t.

Now that you've seen how the integrals work (both forward and backward ), is the IF method a bit clearer?

That last bit cleared it right up.  Thanks for that.  For some reason that x was really throwing me off but you've explained the process much more clearly than either my professor or TA was able to.  If you don't mind me asking, what's your degree in?

Glad it's a little clearer now

I have a mechanical engineering degree.