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Math Help - Solving differential equations using a change of variable

  1. #1
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    Angry Solving differential equations using a change of variable

    (x+y)dy/dx = x^2+xy+x+1 ; y=v-x

    So i did the substitution and ended up with vdv/dx - v =xv+x+1
    and, if correct, i've no idea how to solve.
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  2. #2
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    Quote Originally Posted by Erghhh View Post
    (x+y)dy/dx = x^2+xy+x+1 ; y=v-x

    So i did the substitution and ended up with vdv/dx - v =xv+x+1
    and, if correct, i've no idea how to solve.

    (x+y)dy/dx = x^2+xy+x+1 ; y=v-x

    vdv/dx - v =xv+x+1

    vdv/dx = v+xv+x+1

    vdv/dx = v(1+x) + (1+x)

    vdv/dx = (v+1)(1+x)

    \frac{vdv}{v+1} =(x+1)dx

    Integrate both sides

    \int{\frac{vdv}{v+1}} =\int{(x+1)dx}


    Integrate this thing and enjoy the sunny day
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  3. #3
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    So then i got (y+x)ln|x+y+1|=(1/2)(x^2) +x+c

    The answer, according to my book, is y=ln|x+y+1| +(1/2)(x^2)+C

    I'm unsure of how to get there..
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  4. #4
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    Hi

    A sign - is missing
    (y+x)-ln|x+y+1|=(1/2)(x^2) +x+c

    Then y=ln|x+y+1| +(1/2)(x^2)+c
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  5. #5
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    Can you explain where that minus sign has come from?
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  6. #6
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    \int{\frac{vdv}{v+1}} = \int{\frac{(v+1-1)dv}{v+1}} = \int \left(1-\frac{1}{v+1}\right)\:dv = v - \ln|v+1|
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  7. #7
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    \frac{v}{v+1}= 1- \frac{1}{v+1}
    Intgrate that.

    running-gag got in ahead of me!
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