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Math Help - Homogeneous DE with Transformation

  1. #1
    Senior Member bugatti79's Avatar
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    Homogeneous DE with Transformation

    All,

    I don know how to solve this linear ordinary DE given the transfomation v=x+y. I havent come across this before. I have solved other ones involving y=vx where v is a function of x, thereby usnig the product rule to get dy/dx etc


    <br />
\frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}<br />

    Any clues will be appreciated.
    Thanks
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  2. #2
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    TheEmptySet's Avatar
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    Quote Originally Posted by bugatti79 View Post
    All,

    I don know how to solve this linear ordinary DE given the transfomation v=x+y. I havent come across this before. I have solved other ones involving y=vx where v is a function of x, thereby usnig the product rule to get dy/dx etc


    <br />
\frac{\text{dy}}{\text{dx}}=-\frac{x+y}{3 x+3 y-4}=-\frac{v}{3 v-4}<br />

    Any clues will be appreciated.
    Thanks
    You are almost there but you need to find \frac{dy}{dx} is terms of v.

    \displaystyle v=x+y \implies \frac{dv}{dx}=1+\frac{dy}{dx} \implies \frac{dy}{dx}=\frac{dv}{dx}-1

    Putting this into what you have above gives

    \displaystyle \frac{dv}{dx}-1=-\frac{v}{3v-4}

    can you finish from here?
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    You are almost there but you need to find \frac{dy}{dx} is terms of v.

    \displaystyle v=x+y \implies \frac{dv}{dx}=1+\frac{dy}{dx} \implies \frac{dy}{dx}=\frac{dv}{dx}-1

    Putting this into what you have above gives

    \displaystyle \frac{dv}{dx}-1=-\frac{v}{3v-4}

    can you finish from here?
    ok, from your last line,I rearrange, separate variables and integrate

    \int \frac{3v-4}{2v-4}dv=\int dx from this I get

    \int \frac{3}{2} dv+\int \frac{1}{v-2}dv=\int dx therefore, after some manipulation, giving

    x+3y+2ln((x+y)-2)=-2c

    The answer in the book for the LHS is the same and for the RHS its A, so I could call -2c =A, right?...
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by bugatti79 View Post
    ok, from your last line,I rearrange, separate variables and integrate

    \int \frac{3v-4}{2v-4}dv=\int dx from this I get

    \int \frac{3}{2} dv+\int \frac{1}{v-2}dv=\int dx therefore, after some manipulation, giving

    x+3y+2ln((x+y)-2)=-2c

    The answer in the book for the LHS is the same and for the RHS its A, so I could call -2c =A, right?...
    Yes you can relable your constants.
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  5. #5
    Senior Member bugatti79's Avatar
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    can I retract the double reply...I sent it by accident? I dont want to violate the forum rules?
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