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Math Help - Separation of Variables

  1. #1
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    Separation of Variables

    I have to use sep. of var. to solve the following diff eq:

    e^(t)*y dy/dt = e^(-y) + e^(-2t - y)

    MY WORK:

    e^(t)*y dy = (e^(-y) + e^(-2t - y)) dt

    y dy = (e^(-y) + e^(-2t - y))/(e^(t)) dt

    I'm not sure how to get the y to the LHS.
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  2. #2
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    e^(t)*y dy/dt = e^(-y) + e^(-2t - y)

    ye^t dy/dt = e^(-y) + e^(-(2t +y))
    Put them all in positive exponents,
    ye^t dy/dt = 1/(e^y) + 1/(e^(2t +y))
    ye^t dy/dt = 1/(e^y) + 1/[(e^(2t))(e^y)]
    Make the two fractions in the RHS into one fraction only,
    common denominator is e^(2t) *e^y,
    ye^t dy/dt = [e^(2t) +1] / [(e^(2t))(e^y)]
    Clear the fraction, multiply both sides by [(e^(2t))(e^y)],
    [(e^(2t))(e^y)][ye^t] dy/dt = e^(2t) +1
    (ye^y)(e^(3t)) dy/dt = e^(2t) +1
    (ye^y) dy/dt = [e^(2t) +1] /[e^(3t)]
    (ye^y) dy = ([e^(2t) +1] / [e^(3t)])dt
    (ye^y)dy = {(e^(2t)/(e^(3t)) +1/(e^(3t))}dt
    (ye^y)dy = {1/(e^t) +1/(e^(3t))}dt

    You want to continue from here?
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  3. #3
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    Hello, Ideasman!

    I have to use sep. of var. to solve the following diff eq:
    . . e^t\,y\,\frac{dy}{dt} \: =\: e^{-y} + e^{-2t - y}
    Multiply by e^y\!:\;\;\displaystyle{e^y\,e^t\,y\,\frac{dy}{dt} \:=\:1 + e^{-2t}}

    Multiply by e^{-t}\!:\;\;e^y\,y\,\frac{dy}{dt} \:=\:e^{-t} + e^{-3t}

    And we have: . y\,e^y\,dy \;=\;\left(e^{-t} + e^{-3t}\right)\,dt

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