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Math Help - separable equation 2

  1. #1
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    separable equation 2

    By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

    the solutions as the attachment..
    the problem is, how to convert from..
    dy/dx = (x^2 + y^2) / 2x^2

    to...
    xdv/dx = (v-1)^2 / 2
    ???
    Attached Thumbnails Attached Thumbnails separable equation 2-untitled.bmp  
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  2. #2
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    Instead start by saying y=vx; therefore, dy=xdv+vdx.

    Now take your de and multiple by dx to both sides and you will be left with dy = all that stuff.

    Subtract all that stuff to obtain:

    dy-(blah)dx=0

    From taking the derivative of y=vx, we know dy=xdv+vdx. Substitute that in where you see dy and sub in vx where ever there are xs.
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    still can't understand...
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    I typed up some of the steps in maple and converted to pdf.

    You should be able to finish it from there.
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  5. #5
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    Quote Originally Posted by nameck View Post
    By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

    the solutions as the attachment..
    the problem is, how to convert from..
    dy/dx = (x^2 + y^2) / 2x^2

    to...
    xdv/dx = (v-1)^2 / 2
    ???
    \frac{dy}{dx} = \frac{x^2+y^2}{2x^2} = \frac{x^2}{2x^2} +\frac{y^2}{2x^2}= \frac{1+v^2}{2}

    now since when sub in v we have to find

    v = \frac{y}{x}

    vx =y

    \frac{dv}{dx}x + v = \frac{dy}{dx}

    now put it all together

    \frac{dv}{dx}x + v = \frac{1+v^2}{2}

    \frac{dv}{dx}x = \frac{1+v^2}{2} - v

    \frac{dv}{dx}x = \frac{1+v^2}{2} - \frac{2v}{2}

    \frac{dv}{dx}x = \frac{v^2-2v +1}{2}

    \frac{dv}{dx}x = \frac{(v-1)^2}{2}

    \int\frac{dv}{(v-1)^2} = \int\frac{dx}{2x}
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  6. #6
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    Quote Originally Posted by nameck View Post
    By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

    the solutions as the attachment..
    the problem is, how to convert from..
    dy/dx = (x^2 + y^2) / 2x^2

    to...
    xdv/dx = (v-1)^2 / 2
    ???
    y = xv \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} using the product rule.

    And obviously y = xv \Rightarrow \frac{x^2 + y^2}{2x^2} = \frac{x^2 + x^2 v^2}{2x^2} = ....
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  7. #7
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    If v = \frac{y}{x} then y = vx.

    If you take the derivative of both sides, you get

    \frac{d}{dx}(y) = \frac{d}{dx}(vx)

    \frac{dy}{dx} = v\,\frac{d}{dx}(x) + x\,\frac{d}{dx}(v)

    \frac{dy}{dx} = v + x\,\frac{dv}{dx}.


    Putting this information into your DE:

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

    v + x\,\frac{dv}{dx} = \frac{x^2 + \left(vx\right)^2}{2x^2}

    v + x\,\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x^2}

    v + x\,\frac{dv}{dx} = \frac{1 + v^2}{2}

    x\,\frac{dv}{dx} = \frac{1 + v^2}{2} - v

    x\,\frac{dv}{dx} = \frac{v^2 - 2v + 1}{2}

    x\,\frac{dv}{dx} = \frac{(v- 1)^2}{2}

    \frac{2}{(v - 1)^2}\,\frac{dv}{dx} = \frac{1}{x}

    \int{\frac{2}{(v - 1)^2}\,\frac{dv}{dx}\,dx} = \int{\frac{1}{x}\,dx}

    \int{2(v - 1)^{-2}\,dv} = \int{\frac{1}{x}\,dx}.

    You should be able to go from here.
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