Question

okey this is a tough 1
x^2 * (dy/dx) + xy =1, now to solve this I must use the integrating factor method. but im stuck.

Answer 1

is it solving for dy/dx?? Or what??

Answer 2

Integrating Factor Method : \[\left(\begin{matrix}dy \\dx \end{matrix}\right) + P(x) = Q(x)\]
then the integrating factor is
\[\mu = e \int\limits_{}^{} p(x)dx\]
and y is given by
\[muy = \int\limits_{}^{} \mu Q(x)\]

Answer 3

Divide each side by x^2 to get it into that nice standard form:\[\Large\rm y'+\frac{1}{x}y=\frac{1}{x^2}\]Where is it that you're stuck?

Answer 4

So far I have (dy/dx) + y/x = 1/x^2
which gives p(x) = 1/x and q(x) = 1/x^2
I got (mu) to u= e^ln(x)

Answer 5

so what is the integrate of e^ln(x)* (1/x^2)? there's where I get stuck.

Answer 6

Mmm ok great,\[\Large\rm u=e^{\ln x}\]This is kind of a thing you'll want to remember.
It's going to help out a lot in dealing with differential equations.

Since the exponential base e, and the log base e are inverse operations, they "undo" one another.\[\Large\rm e^{\ln x}=x\]

Answer 7

Great now, theintegrate of 1/x^2, it can be written as ln(x^2) and x^-1 which is same as 1/x so which of the integrate is the correct 1 for 1/x^2?

Answer 8

I didn't quite follow that...
Your integrating factor is simply x.
So after multiplying through by x, your problem can be simplifies by using product rule in reverse,\[\Large\rm (xy)'=x\frac{1}{x^2}\]Integrating gives:\[\Large\rm xy=\int\limits \frac{1}{x}dx\]You remember how to integrate 1/x?

Answer 9

yes I got confused, but I solved it and got the correct answer, the only problem I have is remembering which integrating rule that applies since there is so much integration in this course, btw I want to ask.
Is this calcuulus 1 or 2? because we do everything in english and have exams in swedish.

Answer 10

This material is normally introduced near the end of Calculus 2.
At least that's normally how it goes here in the United States.

Answer 11

okey thanks for the fast reply, I joined a few days ago, I love this site:)

Answer 12

Hehe, me too ^^

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 13

thank you, I'll try to take it from here and see how far I get :)

Answer 14

I think you're going to cause yourself a lot of pain @Xlegalize if you try to memorize any formulas for using an integrating factor.

Understanding the process makes the whole thing much simpler, as you're really just multiplying by some function that you make up so that you can reverse the product rule.

Answer 15

Yes @Kainui it seems you're right, memorizing the formulas has really been a pain. I'll have to read through more thoroully:)