# Thread: First order integrating factors

1. ## First order integrating factors

$\displaystyle ty'+ 5y= t^2 - t + 6$ , with initial:$\displaystyle y(1)=5, t>0$

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.

$\displaystyle y'+p(x)= q(x)$

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

$\displaystyle e^\int 5t$

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

Thanks!

2. ## Re: First order integrating factors

Divide both sides by $\displaystyle \displaystyle t$ then the integrating factor becomes $\displaystyle \displaystyle e^{\int \frac{5}{t}~dt}$

3. ## Re: First order integrating factors

Dividing by t gives me :

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

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

multiply both sides by$\displaystyle 5t$ :

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

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

Did I set up my integration correctly?

Thanks.

4. ## Re: First order integrating factors

Originally Posted by terrygrada
Dividing by t gives me :

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

integrating factor then from $\displaystyle e^/int 5/t dt$ becomes $\displaystyle 5t$
No, it isn't. What is the integral of 5/t? What is the exponential of that?

multiply both sides by$\displaystyle 5t$ :

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

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

Did I set up my integration correctly?

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

Your original equation was

5. ## Re: First order integrating factors

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 ?

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

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???

6. ## 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:

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

We compute the integrating factor:

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

Multiplying the ODE by this factor, we get:

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

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

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

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

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

7. ## Re: First order integrating factors

Originally Posted by terrygrada
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 \displaystyle \begin{align*} \int{\frac{5}{t}\,dt} = 5\ln{t} \end{align*}, so the integrating factor is

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