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Math Help - solve differential equation HELP pls

  1. #1
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    solve differential equation HELP pls

    dy/dx= (2x-y)/(x+y) all subject to y(1) = 1

    I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help
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  2. #2
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    Quote Originally Posted by manalive04 View Post
    dy/dx= (2x-y)/(x+y) all subject to y(1) = 1

    I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help
    That sounds like a very good idea! What did you get?
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  3. #3
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    i got an improper fraction which i am not sure what to do with

    x(dv/dx) = (2 - 2v - v^2)/1+v

    then took everything to the other side ==>

    (1+v)/(2-2v-v^2)dv = dx/x

    then to intergrate both sides

    but how do you do the LHS? it is topheavy?
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  4. #4
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    Quote Originally Posted by manalive04 View Post
    i got an improper fraction which i am not sure what to do with

    x(dv/dx) = (2 - 2v - v^2)/1+v

    then took everything to the other side ==>

    (1+v)/(2-2v-v^2)dv = dx/x

    then to intergrate both sides

    but how do you do the LHS? it is topheavy?
    \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx

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

    can you integrate the LHS now?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx

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

    can you integrate the LHS now?
    can you plese show me how im sorry im not confident here
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  6. #6
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    hint: \int \frac{u'}{u} \, dv
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  7. #7
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    Quote Originally Posted by skeeter View Post
    hint: \int \frac{u'}{u} \, dv
    what is u'

    im sorry i dont understand
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  8. #8
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    Quote Originally Posted by manalive04 View Post
    what is u'

    im sorry i dont understand
    u is just a variable in this case.

    \int{\frac{u'}{u}} is a special case of integral and it equals ln|x| + C

    it is saying that differentiating 2-2v-v^2 (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v
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  9. #9
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    Quote Originally Posted by e^(i*pi) View Post
    u is just a variable in this case.

    \int{\frac{u'}{u}} is a special case of integral and it equals ln|x| + C

    it is saying that differentiating 2-2v-v^2 (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v
    so what does this imply???

    im so confused can u please make this clearer im sorry
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  10. #10
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    \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx

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

    \ln|2 - 2v - v^2| = -2\ln|x| + C
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