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Math Help - First order integrating factors

  1. #1
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    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!
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  2. #2
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    Re: First order integrating factors

    Divide both sides by \displaystyle t then the integrating factor becomes \displaystyle e^{\int \frac{5}{t}~dt}
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    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.
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    Re: First order integrating factors

    Quote Originally Posted by terrygrada View Post
    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?


    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.

    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
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    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???
    Last edited by terrygrada; September 24th 2012 at 07:57 PM.
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    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
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  7. #7
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    Re: First order integrating factors

    Quote Originally Posted by terrygrada View Post
    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*}
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