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Thread: dy/ dx + (y/x) = 2(x^2)

  1. #1
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    dy/ dx + (y/x) = 2(x^2)

    I was told that the differential equation formed form this equation is EXACT. However , i found that the dM / dy not equal to dN/dx .... why ?
    dy/ dx + (y/x) = 2(x^2)-599.jpg
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    Re: dy/ dx + (y/x) = 2(x^2)

    What about the other integrating factor?
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    Re: dy/ dx + (y/x) = 2(x^2)

    Actually, don't you have your $M$ and $N$ transposed?
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    Re: dy/ dx + (y/x) = 2(x^2)

    Quote Originally Posted by Archie View Post
    Actually, don't you have your $M$ and $N$ transposed?
    R(y) is (1/x)( -6x^2) = -6x? , so the integrating factor is e^-6xy ?
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    Re: dy/ dx + (y/x) = 2(x^2)

    No. The equation should be $M \, \mathrm d x + N \, \mathrm d y = 0$, but you have $M \, \mathrm d y + N \, \mathrm d x = 0$.
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    Re: dy/ dx + (y/x) = 2(x^2)

    Quote Originally Posted by xl5899 View Post
    I was told that the differential equation formed form this equation is EXACT. However , i found that the dM / dy not equal to dN/dx .... why ?
    Click image for larger version. 

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    This equation is first order linear, just use an integrating factor...

    $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{1}{x}\,y &= 2\,x^2 \end{align*}$

    Multiply both sides by the integrating factor $\displaystyle \begin{align*} \mathrm{e}^{\int{\frac{1}{x}\,\mathrm{d}x}} = \mathrm{e}^{\ln{(x)}} = x \end{align*}$ and the equation becomes

    $\displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y &= 2\,x^3 \\ \frac{\mathrm{d}}{\mathrm{d}x} \, \left( x\,y \right) &= 2\,x^3 \\ x\,y &= \int{ 2\,x^3\,\mathrm{d}x} \\ x\,y &= \frac{x^4}{2} + C \\ y &= \frac{x^3}{2} + \frac{C}{x} \end{align*}$
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