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Math Help - Help with a basic equation

  1. #1
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    Help with a basic equation

    Hello everyone,
    I need some help with the following ordinary differential equation:
    y'=(2y-x)/(2x-y), and y(2)=3

    and I also have to solve it for y(0)=0 and explain what happens exactly.

    I'll appreciate any help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GIPC View Post
    Hello everyone,
    I need some help with the following ordinary differential equation:
    y'=(2y-x)/(2x-y), and y(2)=3

    and I also have to solve it for y(0)=0 and explain what happens exactly.

    I'll appreciate any help.
    this is homogenious.

    note that you have \displaystyle \frac {dy}{dx} = \frac {2 \frac yx - 1}{2 - \frac yx}

    Let \displaystyle v = \frac yx \implies y = vx \implies \frac {dy}{dx} = x \frac {dv}{dx} + v

    So your equation becomes:

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

    Now that equation is separable. continue...
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  3. #3
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    let me see if I got it right.

    I get
    x*dv=(v^2-1)/(2-v) dx
    I integrate both sides and get
    x*v=[(v^2-1)/(2-v)]*x +C

    using v=y/x
    y=...=[y^2-x^2]/(2x^2-xy) + C

    now what do I do from here?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GIPC View Post
    let me see if I got it right.

    I get
    x*dv=(v^2-1)/(2-v) dx
    I integrate both sides and get
    x*v=[(v^2-1)/(2-v)]*x +C

    using v=y/x
    y=...=[y^2-x^2]/(2x^2-xy) + C

    now what do I do from here?
    you can't integrate like that!

    It is a separable equation, meaning, you have to separate the variables. Only x's should be on one side, and only v's on the other. this includes the dx and dv
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  5. #5
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    how about this then

    we reverse the sides and get
    (2-v)/(v^2-1) dv = dx/x
    integrate and get
    1/2[ln(1-v) - 3ln(v+1)]= lnx +C
    so
    ln[(1-y/x)/(y/x+1)^3]=2lnx+2c
    (1-y/x)/[(y/x+1)^3]=x*e^2c

    what shoud i do from here?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GIPC View Post
    how about this then

    we reverse the sides and get
    (2-v)/(v^2-1) dv = dx/x
    integrate and get
    1/2[ln(1-v) - 3ln(v+1)]= lnx +C
    so
    ln[(1-y/x)/(y/x+1)^3]=2lnx+2c
    (1-y/x)/[(y/x+1)^3]=x*e^2c

    what shoud i do from here?
    Lets take it from \displaystyle \ln \left| \frac {v - 1}{(v + 1)^3} \right| = 2 \ln |x| + C = 2 \ln |Ax| = \ln (Ax)^2

    Then we have

    \displaystyle \frac {v - 1}{(v + 1)^3} = (Ax)^2

    \displaystyle \Rightarrow \frac {\frac yx - 1}{\left( \frac yx + 1 \right)^3} = (Ax)^2

    multiply the left side by \displaystyle \frac {x^3}{x^3} to get

    \displaystyle \frac {x^2y - x^3}{(y + x)^3} = (Ax)^2

    divide both sides by \displaystyle x^2 to get

    \displaystyle \frac {y - x}{(y + x)^3} = A^2

    Plug in your initial data: \displaystyle y(2) = 3 and solve for \displaystyle A. you end up with

    \displaystyle \frac {y - x}{(y + x)^3} = \frac 1{125}

    that's your solution, defined implicitly.
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