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Math Help - Differential Equations Intergating Factors

  1. #1
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    Differential Equations Intergating Factors

    Hi

    i have a few question i am having trouble solving:
    Could someone tell me where my mistakes are.
    1) \frac{dy}{dx}+\frac{y}{x}=y^3

    \frac{dy}{dx}+y=xy^3

    \frac{dy}{dx}-y^{-2}=\frac{-1}{x}

    I=e^{-x}

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

    e^{-x}y=\frac{e^{-x}}{x}

    what should i do next if this is correct.

    2) xcosy\frac{dy}{dx}-siny=0

    someone tell me how to start this?

    i tried to integrate from this:
    \frac{dy}{dx} -tany = \frac{1}{x}

    but this didn't work.
    P.S
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  2. #2
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    With your first question, you managed to make three mistakes in a row. This problem seems to me to be Bernoulli. I think you need to use that procedure before doing the integrating factors method (which I think you WILL need when you get yourself a first-order LINEAR equation).

    With your second equation, again, you're making algebra mistakes.

    You've got x\cos(y)\,y'-\sin(y)=0. Dividing through by \cos(y) yields x\,y'-\tan(y)=0. Then what?
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  3. #3
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    For 2) the DE is separable...

    x\cos{y}\,\frac{dy}{dx} - \sin{y} = 0

    x\,\frac{dy}{dx} - \frac{\sin{y}}{\cos{y}} = 0

    x\,\frac{dy}{dx} = \frac{\sin{y}}{\cos{y}}

    \frac{1}{x}\,\frac{dx}{dy} = \frac{\cos{y}}{\sin{y}}

    \int{\frac{1}{x}\,\frac{dx}{dy}\,dy} = \int{\frac{\cos{y}}{\sin{y}}\,dy}

    \int{\frac{1}{x}\,dx} = \int{\frac{1}{u}\,du} after making the substitution u = \sin{y}

    \ln{|x|} + C_1 = \ln{|u|} + C_2

    \ln{|x|} = \ln{|\sin{y}|} + C where C = C_2 - C_1

    \ln{|x|} -\ln{|\sin{y}|} = C

    \ln{\left|\frac{x}{\sin{y}}\right|} = C

    \left|\frac{x}{\sin{y}}\right| = e^C

    \frac{x}{\sin{y}} = \pm e^C

    \frac{x}{\sin{y}} = A where A = \pm e^C

    x = A\sin{y}

    \frac{1}{A}x = \sin{y}

    Bx = \sin{y} where B = \frac{1}{A}

    y = \arcsin{Bx}.
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  4. #4
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    oh i c, thank you
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  5. #5
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    You're very welcome for whatever I was able to contribute. Have a good one!
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