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Math Help - Differential equation changing variable?

  1. #1
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    Differential equation changing variable?

    I don't know this one:

    \frac{dy}{dt}=(y+t)^2

    Use substitution z=y-t it says, but then \frac{dy}{dt}=z^2? Then what? Please
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  2. #2
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    Quote Originally Posted by TeaWithoutMussolini View Post
    I don't know this one:

    \frac{dy}{dt}=(y+t)^2

    Use substitution z=y-t it says, but then \frac{dy}{dt}=z^2? Then what? Please
    Not quite if

    z=y-t \implies \frac{d}{dt}z =\frac{d}{dt}(y-t) \iff \frac{dz}{dt}=\frac{dy}{dt}-1
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  3. #3
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    I expect you have to actually make the substitution \displaystyle z = y + t.

    Then \displaystyle y = z - t \implies \frac{dy}{dt} = \frac{dz}{dt} - 1.

    Substituting into the DE \displaystyle \frac{dy}{dt} = (y + t)^2 gives

    \displaystyle \begin{align*} \frac{dz}{dt} - 1 &= z^2 \\ \frac{dz}{dt} &= z^2 + 1 \\ \frac{1}{z^2 + 1}\,\frac{dz}{dt} &= 1 \\ \int{\frac{1}{z^2 + 1}\,\frac{dz}{dt}\,dt} &= \int{1\,dt} \\ \int{\frac{1}{z^2 + 1}\,dz} &= t + C_1 \\ \arctan{z} + C_2 &= t + C_1 \\ \arctan{z} &= t + C \textrm{ where }C = C_1 - C_2 \\ z &= \tan{(t + C)} \\ y + t &= \tan{(t + C)} \\ y &= \tan{(t + C)} - t \end{align*}
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    Sorry yes its y=z-t, so yes I rearranged it to z=y+t. Sorry I typed the wrong thing.

    Can you take a 1/2 out of the LHS and put a two on top integrate to (1/2)log(z^2+1)?
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    Quote Originally Posted by TeaWithoutMussolini View Post
    Sorry yes its y=z-t, so yes I rearranged it to z=y+t. Sorry I typed the wrong thing.

    Can you take a 1/2 out of the LHS and put a two on top integrate to (1/2)log(z^2+1)?
    Absolutely not. You can only use the method you have suggested if the derivative of \displaystyle z^2 + 1, i.e. \displaystyle 2z is a factor.

    The integral of \displaystyle \frac{1}{1 + x^2} is \displaystyle \arctan{x}.
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    You see there's me being stupid again thinking d/dz z^2 = 2, of course it isn't! my brain sometimes, seriously :S

    Yeah I get that I just miscalculated what I was thinking of. Thanks for putting me back in my place lol.

    so y(t) = tan(t+c) - t, that is the final solution? that's what I have.
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    Quote Originally Posted by TeaWithoutMussolini View Post
    You see there's me being stupid again thinking d/dz z^2 = 2, of course it isn't! my brain sometimes, seriously :S

    Yeah I get that I just miscalculated what I was thinking of. Thanks for putting me back in my place lol.

    so y(t) = tan(t+c) - t, that is the final solution? that's what I have.
    It's the final answer if you don't have any boundary conditions to find C...
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