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Math Help - First order differential equation

  1. #1
    Member Em Yeu Anh's Avatar
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    Unhappy First order differential equation

    Solve the differential equation  y' - xy^2 = 2xy

    What I have done:

    \frac{dy}{dx}=2xy+xy^2

    At this point I thought it would be best to factor out the x term.

     \frac{dy}{dx}=x(2y+y^2)

    \frac{dy}{2y+y^2}=xdx

    Is my work so far correct? Looking at the left side I don't know what the next step is.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello Em Yeu Anh
    Quote Originally Posted by Em Yeu Anh View Post
    Solve the differential equation  y' - xy^2 = 2xy

    What I have done:

    \frac{dy}{dx}=2xy+xy^2

    At this point I thought it would be best to factor out the x term.

     \frac{dy}{dx}=x(2y+y^2)

    \frac{dy}{2y+y^2}=xdx

    Is my work so far correct? Looking at the left side I don't know what the next step is.
    Looks good to me, so far!

    Now write 2y+y^2 =y(2+y) and then use Partial Fractions. You can then easily integrate the resulting expression.

    Grandad
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello Em Yeu AnhLooks good to me, so far!

    Now write 2y+y^2 =y(2+y) and then use Partial Fractions. You can then easily integrate the resulting expression.

    Grandad
    Thanks for the help.
    Okay I've gotten it down to these steps:

    \frac{ln|y|}{2}-\frac{ln|2+y|}{2}=\frac{x^2}{2}+C

    Assuming I've made no errors in simplifying,

    \frac{y}{2+y} = Ae^{x^2} where A=+ e^C

    And here I don't know how to isolate for y...Is it allright to split up \frac{y}{2+y} into \frac{y}{2}+1 and go from there?
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  4. #4
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    Quote Originally Posted by Em Yeu Anh View Post
    Thanks for the help.
    Okay I've gotten it down to these steps:

    \frac{ln|y|}{2}-\frac{ln|2+y|}{2}=\frac{x^2}{2}+C

    Assuming I've made no errors in simplifying,

    \frac{y}{2+y} = Ae^{x^2} where A=+ e^C

    And here I don't know how to isolate for y...Is it allright to split up \frac{y}{2+y} into \frac{y}{2}+1 and go from there?
    As you probably know, \frac{x}{a+b} \neq \frac{x}{a} + \frac{x}{b}. What you can do, though, is this:

    \frac{y}{2+y} = Ae^{x^2} \Leftrightarrow \frac{1}{Ae^{x^2}} = \frac{2+y}{y} = 1 + \frac{2}{y}

    And now simplify..
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  5. #5
    MHF Contributor
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    Hello Em Yeu Anh
    Quote Originally Posted by Em Yeu Anh View Post
    Thanks for the help.
    Okay I've gotten it down to these steps:

    \frac{ln|y|}{2}-\frac{ln|2+y|}{2}=\frac{x^2}{2}+C

    Assuming I've made no errors in simplifying,

    \frac{y}{2+y} = Ae^{x^2} where A=+ e^C

    And here I don't know how to isolate for y...Is it allright to split up \frac{y}{2+y} into \frac{y}{2}+1 and go from there?
    \frac{y}{2+y} = Ae^{x^2}

    \Rightarrow y= (2+y)Ae^{x^2}
    =2Ae^{x^2}+yAe^{x^2}
    \Rightarrow y(1-Ae^{x^2})=2Ae^{x^2}

    \Rightarrow y=\frac{2Ae^{x^2}}{(1-Ae^{x^2})}

    which you could write as:

    y=\frac{2e^{x^2}}{(A'-e^{x^2})}, replacing \frac1A by A'.

    Grandad
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  6. #6
    MHF Contributor Calculus26's Avatar
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    My 2 cents:

    An eqn like y ' + p(x) y = q(x) y^n

    is an example of a Bernoulli equataion which can easily be solved

    by letting v = y ^ (1-n)



    or y ' -2xy = x y^2

    is an example of a Bernoulli equataion n = 1

    Let v= y ^(-1)

    dv/dx = -y^(-2)dy/dx

    -y^2 dv/dx - 2xy = x y^2

    dv/dx + 2x v = -x

    Which can now be solved with the intergating factor e^(x^2)

    d(ve^(x^2)) = -xe^(x^2)dx

    v = -1/2 + ce^(-x^2)

    y= 1/v = 1/ [-1/2 + ce^(-x^2)]
    Last edited by Calculus26; January 30th 2010 at 12:58 PM.
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