I am having trouble figuring out how to multiply the integrating factor into the

Question

I am having trouble figuring out how to multiply the integrating factor into the equation.
a. The integrating factor in this problem is x^4
b. The equation is: dy/dx+(4/x)y=x^2-1
c. When the integrating factor gets multiplied in the equation then looks like:
(x^4)dy/dx+(4(x^3)y)=x^6-x^4
d. Then I guess they skip a few steps and get the answer:
d/dx((x^4)y)=x^6-x^4

My question is how do they get from c. to d.? I have been trying to understand this for the past few days and all I can see is you forget about the p(x) when you multiply in the integrating factor. Anyway, are there any steps you are supposed to know to get from point c to point d or are you just supposed to know that p(x) disappears and the dy/dx becomes d/dx and gets multiplied by the (intergratingfactor*y)?

Thanks for any help

Explanation / Answer

The point of multiplying in an integrating factor is to take a nasty expression that has no easy antiderivative and make it into something that DOES have an easy antiderivative. It's kind of like when you complete the square -- the number you're inserting into the equation isn't random; it's specifically engineered to give you something you can then simplify. In the case of an integrating factor, you're usually multiplying in something that will give you a nice product rule. Take this problem. You start with:

dy/dx+(4/x)y = x^2-1

The quantity on the left is a mess; it has no easy antiderivative that you could derive to get dy/dx+(4/x)y. But if we multiply the whole equation through by x^4, we still have a valid equation, and look what we get on the left:

(x^4)dy/dx+4(x^3)y = x^6-x^4

The great thing about the expression on the left now is that *it is now the result of a product rule.* Recall that the product rule says that if we take the derivative of uv, we get:

(uv)' = uv' + vu'

This is what we have! The left-hand side of your equation is (x^4)dy/dx+4(x^3)y, which is x^4 times the derivative of y, plus y times the derivative of x^4! It is exactly the result of a product rule! If we're looking at the above formula with u's and v's, in our case u = x^4 and v = y, so u' = du/dx = 4x^3 and v' = dv/dx = dy/dx, and what we have is uv' + vu'. So this equals (uv)', or ((x^4)y)', which we can write as

d/dx ((x^4)y).

So we can replace the left-hand side of the equation with this and get

d/dx ((x^4)y) = x^6-x^4,

which is step (d).

So to answer your questions you wrote at the end -- I'm not sure what you mean by p(x); that must be notation your book uses for one of the quantities in the equation -- but the answer is no, it is not just a form that you flip to after you multiply in an integrating factor; you are multiplying the integrating factor in to get something that is *specifically* in the form of the derivative of a product, and you can therefore write it as the derivative of the product, but with the product condensed together before the product rule has been applied. Recall that d/dx is an operator -- it's the differential operator that says "I'm deriving in terms of x." So it's not that dy/dx becomes d/dx or anything -- dy/dx is d/dx *applied to y*, or (d/dx) (y), which we can write dy/dx. In the equation, dy/dx means the derivative of y in terms of x. Once we determine that we have the result of a product rule, we can rewrite the expression as the derivative of a product, which we can write as d/dx of the whole product.

Hopefully this answered your question. If you are still having trouble, look at the problem backward. In (d), you have:

d/dx ((x^4)y) = x^6-x^4

which means the left side is "the derivative of (x^4)y in terms of x." Take the derivative of (x^4)y in terms of x -- you'll need the product rule. The first term, x^4, times the derivative of y, plus the second term, y, times the derivative of x^4, is:

d/dx ((x^4)y) = (x^4)(dy/dx) + (y)(4x^3)

which is exactly the expression in (c), so it is clear that you can also turn (x^4)(dy/dx) + (y)(4x^3) back into d/dx ((x^4)y) when going forward from (c) to (d). Hopefully that can help prove it to yourself that this works.

To summarize: All the step you are asking about is is the product rule going backwards. All the rules for how to find what integrating factor to multiply by are specifically formulated so that you always get the result of a product rule and can go backwards to rewrite it as the derivative of a product. You can always check you've done an integrating factor step right by applying the product rule after you condense the expression (like we just did going from (d) back to (c)) and making sure you get what you had when you first multiplied in the integrating factor.

Hope that helps!

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