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Math Help - Linear first-order equation Problem

  1. #1
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    Question Linear first-order equation Problem

    I am quite not sure about my solution to solve this by using Linear equation method.

    y\frac{dx}{dy}-x = 2y^2 ; y(1) = 5

    Then, to follow the form \frac{dy}{dx}+P(x)y=f(x) I get \frac{dx}{dy}-(\frac{1}{y})x=2y (Because I assume, x=y and y=x)
    Integrating factor is e^{\int\frac{1}{y}dy} = e^{ln|y|} = y
    After that, I solved:
    \frac{d}{dy}(xy) = 2y^2
    xy = \frac{2y^3}{3}+C1
    x = \frac{2y^2}{3}+C2

    After I have done like this, I don't know what should I do with y(1) = 5.

    Please help me about this. I am new to Differential Equation.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by noppawit View Post
    I am quite not sure about my solution to solve this by using Linear equation method.

    y\frac{dx}{dy}-x = 2y^2 ; y(1) = 5

    Then, to follow the form \frac{dy}{dx}+P(x)y=f(x) I get \frac{dx}{dy}-(\frac{1}{y})x=2y (Because I assume, x=y and y=x)
    Integrating factor is e^{\int\frac{1}{y}dy} = e^{ln|y|} = y
    After that, I solved:
    \frac{d}{dy}(xy) = 2y^2
    xy = \frac{2y^3}{3}+C1
    x = \frac{2y^2}{3}+C2
    you have the right idea, however, you should have realized that your integrating factor is off. \frac d{dy}(xy) does not yield y \frac {dx}{dy} - x as it should. rather your integrating factor is given by e^{\int - \frac 1ydy}

    After I have done like this, I don't know what should I do with y(1) = 5.
    y(1) = 5 simply means when x = 1, y = 5. you are supposed to plug in these values into your solution (after you correct it ) and this will allow you to solve for the arbitrary constant C_2
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