# First order integrating factors

• Sep 24th 2012, 04:48 PM
First order integrating factors
$ty'+ 5y= t^2 - t + 6$ , with initial: $y(1)=5, t>0$

How do I find the integrating factor from this equation?

I am not sure if the right side follows the standard form but it doesn't look like it.

$y'+p(x)= q(x)" alt="y'+p(x)= q(x)" />

If it does follow the form, I think that the integrating factor is

$e^\int 5t" alt="e^\int 5t" />

Also how do I change so that it doesn't say "font color"?

Thanks!
• Sep 24th 2012, 05:00 PM
pickslides
Re: First order integrating factors
Divide both sides by $\displaystyle t$ then the integrating factor becomes $\displaystyle e^{\int \frac{5}{t}~dt}$
• Sep 24th 2012, 05:19 PM
Re: First order integrating factors
Dividing by t gives me :

$ty'/t +5y/t=t+6$

integrating factor then from $e^/int 5/t dt$ becomes $5t$

multiply both sides by $5t$ :

$(5t)tdy/t+5y/t=t+6(5t) dt$

$\int 5t+5y/t = \int 5t^2 + 30t dt$

Did I set up my integration correctly?

Thanks.
• Sep 24th 2012, 05:35 PM
HallsofIvy
Re: First order integrating factors
Quote:

Dividing by t gives me :

$ty'/t +5y/t=t+6$

integrating factor then from $e^/int 5/t dt$ becomes $5t$

No, it isn't. What is the integral of 5/t? What is the exponential of that?

Quote:

multiply both sides by $5t$ :

$(5t)tdy/t+5y/t=t+6(5t) dt$

$\int 5t+5y/t = \int 5t^2 + 30t dt$
What happened to "dy"?? And you should understand that you cannot integrate a function of y and t with respect to y.

Quote:

Did I set up my integration correctly?

Thanks.
No, you haven't. Are you clear on what the point of an integrating factor is?

• Sep 24th 2012, 05:55 PM
Re: First order integrating factors
Quote:

No, it isn't. What is the integral of 5/t? What is the exponential of that?
integral of 5/t is ln5t and the exponent of ln5t is 5t ? Sorry my mistake

Should be exp^[5 *ln(t)] which = t ?

Quote:

What happened to "dy"?? And you should understand that you cannot integrate a function of y and t with respect to y.
Forgot to include the dy in here

Quote:

No, you haven't. Are you clear on what the point of an integrating factor is?
I don't know much about integrating factors, but I do know it is to help make an equation that would normally be unsolvable, solvable. I would like to learn more about them and how to solve equations like these.

I cannot integrate a function of y and t with respect to y. Do I separate all the variables, y on the left and t on the right? But then wouldn't the integrating factor still bring me a function of t on both sides???
• Sep 24th 2012, 07:01 PM
MarkFL
Re: First order integrating factors
The point of an integrating factor is to make the left side of the linear equation the product of a differentiation.

Given a linear equation in standard form:

$\frac{dy}{dx}+P(x)y=Q(x)$

We compute the integrating factor:

$\mu(x)=e^{\int P(x)\,dx}$

Multiplying the ODE by this factor, we get:

$e^{\int P(x)\,dx}\frac{dy}{dx}+P(x)e^{\int P(x)\,dx}y=e^{\int P(x)\,dx}Q(x)$

$\frac{d}{dx}\left(e^{\int P(x)\,dx}y \right)=e^{\int P(x)\,dx}Q(x)$

$\int\,d\left(e^{\int P(x)\,dx}y \right)=\int e^{\int P(x)\,dx}Q(x)\,dx$

$e^{\int P(x)\,dx}y=\int e^{\int P(x)\,dx}Q(x)\,dx$

$y(x)=e^{-\int P(x)\,dx}\int e^{\int P(x)\,dx}Q(x)\.dx$
• Sep 24th 2012, 07:20 PM
Prove It
Re: First order integrating factors
Quote:

integral of 5/t is ln5t and the exponent of ln5t is 5t ? Sorry my mistake

Should be exp^[5 *ln(t)] which = t ?

Forgot to include the dy in here

I don't know much about integrating factors, but I do know it is to help make an equation that would normally be unsolvable, solvable. I would like to learn more about them and how to solve equations like these.

I cannot integrate a function of y and t with respect to y. Do I separate all the variables, y on the left and t on the right? But then wouldn't the integrating factor still bring me a function of t on both sides???

No, \displaystyle \begin{align*} \int{\frac{5}{t}\,dt} = 5\ln{t} \end{align*}, so the integrating factor is

\displaystyle \begin{align*} e^{\int{\frac{5}{t}\,dt}} = e^{5\ln{t}} = e^{\ln{\left(t^5\right)}} = t^5 \end{align*}