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Math Help - Separating the variables or some other way?

  1. #1
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    Separating the variables or some other way?

    Hello all,

    I'm being confused by a differential equation problem. Solve:

    (2x + 4x^2) dy/dx = e^y

    My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

    Any help appreciated.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by DangerousDave View Post
    Hello all,

    I'm being confused by a differential equation problem. Solve:

    (2x + 4x^2) dy/dx = e^y

    My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

    Any help appreciated.
    (2x + 4x^2) \frac{dy}{dx} = e^y

     e^{-y} dy = \frac{1}{x(2+4x)} dx

    Your original conclusion that this is a seperable equation and that you must use partial fractions on the R.H.S is correct.

    Let's just work through it.

    Partial Fractions,

     \frac{A}{x} + \frac{B}{2+4x} = \frac{A(2+4x) + B(x)}{x(2+4x)}

    2A = 1

     4A + B = 0

    A= \frac{1}{2} and  B = -2

    So the R.H.S becomes

     \frac{1}{2} \int \frac{1}{x} dx - 2 \int \frac{1}{2+4x}dx

     \frac{1}{2} lnx - \frac{1}{2} ln(2+4x) + C

    So far so good.

    Now the L.H.S

    \int e^{-y} dy = - e^{-y}

    L.H.S = R.H.S

    -e^{-y}=\frac{1}{2} lnx - \frac{1}{2} ln(2+4x) + C

    e^{-y}=\frac{1}{2} ln(2+4x) - \frac{1}{2} lnx - C

    Clearly to bring down Y we would use ln as you suggested. Is there any problem with having (as an example)

     ln ( \frac{1}{2} ln(2+4x) )

    Well, let us think about this. What is

    \frac{1}{2} ln(2+4x)

    Well, this will evaluate to a number! So essentially, we'll have ln(some number) which is clearly no problem. So to answer your question, this is not a problem.

    But you should note for

     ln ( \frac{1}{2} ln(2+4x) )

    We have the restriction that,

    \frac{1}{2} ln(2+4x) \ge 1

    Other then that, we are fine!
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  3. #3
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    Quote Originally Posted by DangerousDave View Post
    Hello all,

    I'm being confused by a differential equation problem. Solve:

    (2x + 4x^2) dy/dx = e^y

    My approach has been to transfer the function of x to the right i.e. 1/it and then the e^y to the right yielding e^-y, and then the usual separating the variables method with partial fractions to integrate the right hand side. However, to find y, I ultimately end up with the log of a log, and I was wondering if this is bad news, and whether I should perhaps be using an entirely different method.

    Any help appreciated.
    \frac{dx}{2x+4x^2}=\frac{dy}{e^Y}\Rightarrow

    \int\frac{dx}{2x+4x^2}=\int\frac{dy}{e^y}

    By partial fractions, we obtain:

    \frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{1+2x}dx=-e^{-y}

    Try finishing from here.
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dwsmith View Post
    \frac{dx}{2x+4x^2}=\frac{dy}{e^Y}\Rightarrow

    \int\frac{dx}{2x+4x^2}=\int\frac{dy}{e^y}

    By partial fractions, we obtain:

    \frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{1+2x}dx=-e^{-y}

    Try finishing from here.
    He can solve the problem. He wanted to know if there's any implication of having

    ln(ln(something))

    Which is what you'll get when you finish this problem.
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  5. #5
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    Many thanks for your help. Is it definitely going to be a log of a log in the answer then? There is no alternative method that will produce an alternative neater formulation? (I only ask because these problems usually seem to pan out a bit neater, and when it's messy(ish) then I start to think I did it wrong)
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