Question

I'm reviewing solving linear, first order ODE's using an integrating factor at the moment and am running into trouble doing the problem in full; posted below shortly.

Answer 1

\[xy'-y=x^3+1\]

Answer 2

The trouble that I'm having is that I either have forgotten how to use an integrating factor properly (entirely possible) or that I'm making an error in computing it. Workings so far posted below; I understand that these steps can be skipped, but I'd rather not just use the formula and work through it when practicing.

Answer 3

First, getting the ODE in the appropriate form to use an integrating factor:
\[\frac{dy}{dx}-\frac{y}{x}=\frac{x^3+1}{x}\]

Answer 4

From here, multiplying both sides by our mystery integrating factor, mu:
\[\mu \frac{dy}{dx}-\mu \frac{y}{x}=\mu \frac{x^3+1}{x}\]

Answer 5

\[\mu ' (x)=\mu (x) P(x)\] by definition for this to be true; solving for mu in the typical way, where it yields: \[\mu = e^{\int\limits_{}^{}P(x)dx}\]

Answer 6

\[\mu = e^{\int\limits_{}^{}-\frac{1}{x}dx}=e^{-\ln(x)}=e^{\ln(x^{-1})}=\frac{1}{x}\]

Answer 7

Here's the problem, though, when I take this value and plug it in and differentiate mu*y, I don't get my original problem.

Answer 8

oh okay

Answer 9

so that means you need to plug in something else

Answer 10

Because, for the problem to work, \[\frac{d}{dx}(\mu y)=\mu \frac{dy}{dx}+\mu' y\]

Well...like what? Where did I make a mistake, and how?

Answer 11

okay ley me check it okay

Answer 12

\[\frac{d}{dx}\bigg(\frac{y}{x}\bigg)=\frac{dy}{dx}\frac{1}{x}-\frac{1}{x^2}y\]
I'm getting the original thing in form, divided by a single x.

Answer 13

okay

Answer 14

how did you get you answer so i may known where you went wrong

Answer 15

Well, I got mu doing the typical integrating factor process, outlined above (does that make sense?) and then I differentiated mu times y, which should yield the original equation in that form where y' has a coefficient of mu.

The original equation, when put in form, should just be the result of the product rule for differentiation, between mu and y.

Answer 16

hmmm

Answer 17

do you have answer chioceS?

Answer 18

Shouldn't it look like this after multiplying by your integrating factor?
\[\frac{d}{dx}(\frac{1}{x}y)= \frac{1}{x} \frac{x^3 +1}{x}\]

Answer 19

It's not multiple choice, I'm given the answer but regardless, this is prior to the final answer; I need to clear up this conceptual misunderstanding prior to anything else, because either my integrating factor is wrong or the process itself is wrong;

Yup, that's what the RHS should look like, I'm confused about the LHS

Answer 20

I feel like this is something really simple that I'm missing.

Answer 21

what do you feel like thats missing

Answer 22

Lol, I don't know. @Kainui , could you help me out on this?

Answer 23

displayerror you right:3

Answer 24

Nevermind, I'm an idiot, I think I got it.

Answer 25

:D

Answer 26

your not a idiot

Answer 27

Just forgot how to properly use an integrating factor for a moment, either way, still need to finish the problem in its entirety.

Answer 28

your smart

Answer 29

:D yes

Answer 30

\[\frac{d}{dx}\bigg(\mu (x)y(x)\bigg)=\mu(x)f(x); \ \ \ y(x)=\frac{\int\limits_{}^{}\mu(x)f(x)}{\mu(x)}\]

Answer 31

:D

Answer 32

\[\frac{ dy }{ dx }+Py=Q,\]
where P and Q are functions of x has an integrating factor
\[I.F.=e ^{\int\limits P dx}\]
and complete sol. is
\[y*I.F=\int\limits Q*I.F. dx+c\]

Answer 33

\[y=\frac{\int\limits_{}^{}\frac{x^3+1}{x^2}}{\frac{1}{x}}\]