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    I'm totally stuck on part ii of this question.

    Q i) Solve  \frac{dy}{dt} = y^3 - y,
    y = \frac{1}{\sqrt2} when  t = 0 .

    Answer is:
     y = \frac{1}{\sqrt {1 + e^{2t}}}

    Q ii) solve for y again using subsitution  y = z^{-\frac{1}{2}}, where z = z(t) is a new dependent variable.

    This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

    Could someone solve this please?
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    (Original post by BAD AT MATHS)
    I'm totally stuck on part ii of this question.

    Q i) Solve  \frac{dy}{dt} = y^3 - y,
    y = \frac{1}{\sqrt2} when  t = 0 .

    Answer is:
     y = \frac{1}{\sqrt {1 + e^{2t}}}

    Q ii) solve for y again using subsitution  y = z^{-\frac{1}{2}}, where z = z(t) is a new dependent variable.

    This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

    Could someone solve this please?
    y=z^{-1/2}

    Differentiate to get \frac{dy}{dt}=-\frac{1}{2}z^{-3/2}\frac{dz}{dt}.

    y^3=z^{-3/2}.

    Rewrite the D.E in terms of z and t.
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    (Original post by BAD AT MATHS)
    I'm totally stuck on part ii of this question.

    Q i) Solve  \frac{dy}{dt} = y^3 - y,
    y = \frac{1}{\sqrt2} when  t = 0 .

    Answer is:
     y = \frac{1}{\sqrt {1 + e^{2t}}}

    Q ii) solve for y again using subsitution  y = z^{-\frac{1}{2}}, where z = z(t) is a new dependent variable.

    This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

    Could someone solve this please?
    y=z^{-\frac{1}{2}} \to  \frac{dy}{dt}=-\frac{1}{2}z^{-\frac{3}{2}} \frac{dz}{dt}
    Then substitute and separate the variables
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    (Original post by brianeverit)
    y=z^{-\frac{1}{2}} \to  \frac{dy}{dt}=-\frac{1}{2}z^{-\frac{3}{2}} \frac{dz}{dt}
    Then substitute and separate the variables
    (Original post by BabyMaths)
    y=z^{-1/2}

    Differentiate to get \frac{dy}{dt}=-\frac{1}{2}z^{-3/2}\frac{dz}{dt}.

    y^3=z^{-3/2}.

    Rewrite the D.E in terms of z and t.
    Ok, I see now thanks.

    EDIT: babymaths i tried +ve rep'ing you but I got a message saying "Please rate some other members before rating this member again.".
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    (Original post by BAD AT MATHS)
    Ok, I see now thanks.

    EDIT: babymaths i tried +ve rep'ing you but I got a message saying "Please rate some other members before rating this member again.".
    It really is the thought that counts. Thanks.
 
 
 
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