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Math Help - Differential Equation : y/x substitution

  1. #1
    Member pberardi's Avatar
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    Differential Equation : y/x substitution

    Hey all,
    I used a v = y/x substitution on this and I got a really weird looking answer so I just want to see how right or wrong I am.

    Solve: (x^2 + 2y^2)dx + xydx = 0
    v = y/x and dy/dx = v +x(dv/dx)
    So: x(dv/dx) = (-1/v) - v
    after integrating I get:
    (-1/2)ln|1 + v^2| = ln(x) + C
    subbing back gives
    (1 + y^2/x^2)^2 = x + e^c

    Can someone please check this work and then come up with an equation for y? I have an answer but it looks so weird I don't think its right. Thanks.
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    Quote Originally Posted by pberardi View Post
    Hey all,
    I used a v = y/x substitution on this and I got a really weird looking answer so I just want to see how right or wrong I am.

    Solve: (x^2 + 2y^2)dx + xydx = 0 should this be a dy
    v = y/x and dy/dx = v +x(dv/dx)
    So: v + x(dv/dx) = (-1/v) - 2 v
    after integrating I get:
    (-1/2)ln|1 + v^2| = ln(x) + C
    subbing back gives
    (1 + y^2/x^2)^2 = x + e^c
    should be (-1/2) and *
    Can someone please check this work and then come up with an equation for y? I have an answer but it looks so weird I don't think its right. Thanks.
    A couple of things. First, I think your missing some things - see the red above (there are some other mistakes after that also - in blue). Second, if you write out the equation like

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

    you'll also notice it's Bernoulli.
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  3. #3
    Member pberardi's Avatar
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    (y/x) sub difEq corrections made...please help me finish?

    Thanks Danny,
    I have seen the errors and redid it. If you don't mind I would like to stay with the y/x sub instead of the Bernoulli sub since we have not yet covered them ok?

    Solve: (x^2 + 2y^2)dx + xydy = 0
    xydy = -(x^2 + 2y^2)dx
    dy/dx = -(x/y) - (2y/x)

    let v = y/x, y = vx, dy/dx = v + xdv/dx

    v + x(dv/dx) = -(1/v) - 2v
    please check algebra here:
    x(dv/dx) = (-1-3v^2)/v
    -vdv/(1+3v^2) = dx/x

    with a u substitution I get
    LHS: -(1/6)ln|1 + 3y^2/x^2|
    So:
    ln|1 + 3y^2/x^2| = -6ln|x| + C

    Is this correct so far? Also please could you show me some skills on how to wrap this up correctly. Thanks so much.
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    Quote Originally Posted by pberardi View Post
    Thanks Danny,
    I have seen the errors and redid it. If you don't mind I would like to stay with the y/x sub instead of the Bernoulli sub since we have not yet covered them ok?

    Solve: (x^2 + 2y^2)dx + xydy = 0
    xydy = -(x^2 + 2y^2)dx
    dy/dx = -(x/y) - (2y/x)

    let v = y/x, y = vx, dy/dx = v + xdv/dx

    v + x(dv/dx) = -(1/v) - 2v
    please check algebra here:
    x(dv/dx) = (-1-3v^2)/v
    -vdv/(1+3v^2) = dx/x

    with a u substitution I get
    LHS: -(1/6)ln|1 + 3y^2/x^2|
    So:
    ln|1 + 3y^2/x^2| = -6ln|x| + C

    Is this correct so far? Also please could you show me some skills on how to wrap this up correctly. Thanks so much.
    Yes, then exponentiate

    1 + \frac{3 y^2}{x^2} = e^C e^{-6\ ln|x|} = \frac{c}{x^6}

    I've used the factor that e^C is a constant so I called it c only and

    e^{-6\ ln|x|} = e^{\ln |x|^{-6}} = x^{-6}
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  5. #5
    Member pberardi's Avatar
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    (y/x) sub wrap up

    I have this for an answer, this is the process. How right or wrong is this?

    ln|1 + 3y^2/x^2| = -6ln|x| + C
    1 + 3y^2/x^2 = x^-6 + e^c
    3y^2/x^2 = 1/x^6 + e^c -1
    3y^2 = 1/x^4 + x^2e^c - x^2
    y = sqrt[ (1/3x^4) + (x^2e^c/3) - (x^2/3) ]

    Yes, no, maybe?
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    Quote Originally Posted by pberardi View Post
    I have this for an answer, this is the process. How right or wrong is this?

    ln|1 + 3y^2/x^2| = -6ln|x| + C
    1 + 3y^2/x^2 = x^-6 + e^c (should be x not +)
    3y^2/x^2 = 1/x^6 + e^c -1
    3y^2 = 1/x^4 + x^2e^c - x^2
    y = sqrt[ (1/3x^4) + (x^2e^c/3) - (x^2/3) ]

    Yes, no, maybe?
    See in red. It will change things below that.
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  7. #7
    Member pberardi's Avatar
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    y/x sub wrap up

    Danny,
    I am sorry but I really need some instruction on how to wrap this up. I am a little shaky when it comes to the exponentiation part. Can I see how you would finish this off and get an equation for y that is correct please?
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  8. #8
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    Quote Originally Posted by pberardi View Post
    Danny,
    I am sorry but I really need some instruction on how to wrap this up. I am a little shaky when it comes to the exponentiation part. Can I see how you would finish this off and get an equation for y that is correct please?
    Sure.

    Quote Originally Posted by pberardi View Post
    I have this for an answer, this is the process. How right or wrong is this?

    (1) \ln|1 + 3y^2/x^2| = -6 ln|x| + C
    (2) 1 + 3y^2/x^2 = x^-6 + e^c
    (3) 3y^2/x^2 = 1/x^6 + e^c -1
    (4) 3y^2 = 1/x^4 + x^2e^c - x^2
    (5) y = sqrt[ (1/3x^4) + (x^2e^c/3) - (x^2/3) ]

    Yes, no, maybe?
    From your steps above

    <br />
(1)\;\;\; \ln \left|1 + \frac{3y^2}{x^2} \right| = -6\ln|x| + C<br />

    <br />
(2)\;\;\;1 + \frac{3y^2}{x^2} = x^{-6} e^c\;\;\; (\text{let}\;\; e^c = k)<br />

    <br />
(3)\;\;\;\frac{3y^2}{x^2} = \frac{k}{x^6} -1<br />

    <br />
(4)\;\;\;3y^2 = \frac{k}{x^4} - x^2<br />

    <br />
(5)\;\;\;y = \pm \sqrt{ \frac{k}{x^4} - \frac{x^2}{3} }\;\;\;( \text{the 3 is absorbed into \it{k}})<br />

    Hope it helps.
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    Hello, pberardi!

    I came up with a different answer.
    My journey was quite intricate, but the answer came out rather "neat."


    (x^2 + 2y^2)\,dy + xy\,dx \:= \:0

    We have: . \frac{dy}{dx} \:=\:-\frac{xy}{x^2+y^2}

    Divide top and bottom of the fraction by x^2\!:\;\;\frac{dy}{dx} \:=\:-\frac{\frac{y}{x}}{1 + \left(\frac{y}{x}\right)^2}

    Let: . v = \frac{y}{x}\quad\Rightarrow\quad y \,=\,vx \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}

    Substitute: . v + x\frac{dv}{dx} \;=\;-\frac{v}{1+v^2}

    . . Then: . x\frac{dv}{dx} \;=\;-\frac{v}{1+v^2} - v \;=\;-\frac{v(v^2+2)}{v^2+1}

    Separate variables: . \frac{v^2+1}{v(v^2+2)}\,dv \;=\;-\frac{dx}{x}

    The left side requires Partial Fractions: . \frac{1}{2}\left[\frac{1}{v} + \frac{v}{v^2+2}\right]\,dv \;=\;-\frac{dx}{x}

    Integrate: . \tfrac{1}{2}\left[\ln v + \tfrac{1}{2}\ln(v^2+2)\right] \;=\; -\ln x  + c_1

    Multiply by 2: . \ln v + \ln\sqrt{v^2+1} \;=\;-2\ln x + c_2

    . . . . . \ln\left(v\sqrt{v^2+1}\right) \;=\;\ln\left(x^{-2}\right) + \ln(c_3) \;=\;\ln\left(c_3x^{-2}\right)

    Exponentiate: . v\sqrt{v^2+1} \;=\;\frac{c_3}{x^2}


    Back-substitute: . \frac{y}{x}\sqrt{\frac{y^2}{x^2} + 1} \;=\;\frac{c_3}{x^2} \quad\Rightarrow\quad \frac{y}{x}\,\frac{\sqrt{y^2+x^2}}{x} \;=\;\frac{c_3}{x^2}


    Therefore: . y\sqrt{x^2+y^2} \;=\;c_3 \quad\Rightarrow\quad y^2(x^2+y^2) \;=\;C

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  10. #10
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    Quote Originally Posted by pberardi View Post
    Hey all,
    I used a v = y/x substitution on this and I got a really weird looking answer so I just want to see how right or wrong I am.

    Solve: (x^2 + 2y^2)dx + xydx = 0
    From his original post

    Quote Originally Posted by Soroban View Post
    Hello, pberardi!

    I came up with a different answer.
    My journey was quite intricate, but the answer came out rather "neat."


    We have: . \frac{dy}{dx} \:=\:-\frac{xy}{x^2+y^2}

    I guess it depends what dx you change

    (also, theres a 2 in the original post)
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  11. #11
    Member pberardi's Avatar
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    thank you

    Thanks Danny for all the effort. And thank you too Soroban for chiming in. I really appreciate it.
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