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Math Help - Help with Integrating Factor

  1. #1
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    Help with Integrating Factor

    Can anyone help me solve this equation using the method differentiation with integrating factor:

    dy/dt=(y-1)t^2, where inital condition is y(1)=2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sazzy3 View Post
    Can anyone help me solve this equation using the method differentiation with integrating factor:

    dy/dt=(y-1)t^2, where inital condition is y(1)=2
    this has nothing to do with integrating factors. the equation is separable

    it can be written as:

    \frac {dy}{y - 1} = t^2~dt

    now integrate both sides and continue
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  3. #3
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    I have to solve that equation using integrating factor method

    I have to solve this equation using the method of integrating factor.
    And i have no idea how
    i know i have to change the equation to the form to linear
    A1(t)dy/dt + A0(t)y=f(t)
    then write it as: dy/dt + p(t)y=q(t)
    where p(t)=A0(t)/A1(t) and q(t)=f(t)/A1(t)

    Then solve by finding an integrating factor etc
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sazzy3 View Post
    I have to solve this equation using the method of integrating factor.
    And i have no idea how
    i know i have to change the equation to the form to linear
    A1(t)dy/dt + A0(t)y=f(t)
    then write it as: dy/dt + p(t)y=q(t)
    where p(t)=A0(t)/A1(t) and q(t)=f(t)/A1(t)

    Then solve by finding an integrating factor etc
    fine, it is overkill though

    you have y' = (y - 1)t^2 = yt^2 - t^2

    \Rightarrow y' - t^2y = -t^2

    now you have your p(t) = -t^2. the integrating factor is given by e^{\int p(t)}

    now see post #2 here

    you may also want to see post #21 here
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  5. #5
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    continued

    So i get from
    having alpha(t)= e^ integral from t to t0 -t^2 dt

    then
    alpha(t)= exp [integral from t to t0 -t^2 dt]

    then exp -t^3/3=-t^2
    then

    -t^2dy/dt -t^2y=-t^2 x -t^2


    is this correct so far i am now lost as to where to go from here?
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  6. #6
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    Quote Originally Posted by sazzy3 View Post
    Can anyone help me solve this equation using the method differentiation with integrating factor:

    dy/dt=(y-1)t^2, where inital condition is y(1)=2
    Write the equation as \frac{dy}{dt}- t^2 y= -t^2. An "integrating factor" is a function u(t) such that mutiplying the entire equation by it:
    u(t)\frac{dy}{dt}- t^2u(t)y= -t^2u(t) will make the left side an "exact" derivative: \frac{d(u(t)y)}{dt}.

    By the product rule, \frac{d(u(t)y)}{dt}= u(t)\frac{dy}{dt}+ \frac{du}{dt}y and we want that equal to u(t)\frac{dy}{dt}- t^2u(t)y. That means we must have \frac{du}{dt}= -t^2u which is a separable equation for u: \frac{du}{u}= -t^2dt. Integrating both sides, ln(u)= -\frac{1}{3}t^3 (since we are only looking for a single solution we can ignore the constant of integration) or u(t)= e^{-\frac{t^3}{3}}.

    Multiplying the original equation by that gives e^{-\frac{t^3}{3}}\frac{dy}{dt}- t^2e^{-\frac{t^3}{3}}y= -t^2e^{-\frac{t^3}{3}}.
    Of course, the left side of that is a single derivative:
    \frac{d(e^{-\frac{t^3}{3}})y}{dt}= -t^2e^{-\frac{t^3}{3}}
    Integrate both sides of that (The left side is easy. Use the substitution v= \frac{t^3}{3} on the right.) and solve for y.
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  7. #7
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    Quote Originally Posted by Jhevon View Post

    now see post #2 here

    you may also want to see post #21 here
    And I suggest you to put those links into your signature.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    And I suggest you to put those links into your signature.
    Haha, yes. you have been telling me that for quite some time now. I will put the links I refer people to often in my signature. Save me the trouble of searching for the posts every time!
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