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Math Help - Reducing to a linear equation

  1. #1
    Super Member Deadstar's Avatar
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    Reducing to a linear equation

    This is part of a larger question but I'm stick on this part.

    I need to change \frac{dy}{dt} + f(t)y = g(t)y^a (*) where f \textrm{ and } g are continuous functions and a \in \mathbb{R} into...

    \frac{dz}{dt} + (1-a)f(t)z = (1-a)g(t) using z = y^{1-a}.

    So what I've done so far is...

    Multiply both sides of (*) by y^{-a}
    y^{-a}\frac{dy}{dt} + f(t)y^{1-a} = g(t)

    Multiply both sides by 1-a
    (1-a)y^{-a}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)

    So we now have \frac{dz}{dt}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t) (i think...)

    But how do I get rid of the \frac{dy}{dt}?
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by Deadstar View Post
    This is part of a larger question but I'm stick on this part.

    I need to change \frac{dy}{dt} + f(t)y = g(t)y^a (*) where f \textrm{ and } g are continuous functions and a \in \mathbb{R} into...

    \frac{dz}{dt} + (1-a)f(t)z = (1-a)g(t) using z = y^{1-a}.

    So what I've done so far is...

    Multiply both sides of (*) by y^{-a}
    y^{-a}\frac{dy}{dt} + f(t)y^{1-a} = g(t)

    Multiply both sides by 1-a
    (1-a)y^{-a}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t)

    So we now have \frac{dz}{dt}\frac{dy}{dt} + (1-a)f(t)z = (1-a)g(t) (i think...)

    But how do I get rid of the \frac{dy}{dt}?
    If z = y^{1-a} then wouldn't \frac{dz}{dt} = (1-a)y^{-a} \frac{dy}{dt} ?
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  3. #3
    Super Member Deadstar's Avatar
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    Thanks that's kinda what I thought but wasn't sure.
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