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Math Help - Differential equation - substitution

  1. #1
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    Post Differential equation - substitution

    find the general solution (x+y)y' = (3y-x), where y(1)=2

    I did the substitution y=ux

    I solved to get: ln|v-1| - 2/(v-1) = -ln|x| + A

    I then rearranged and introduced u=y/x back into the equation:

    ln|y-x| - 2x/(y-x) = A

    For y i get

    y = Ce^(2x/(y-x)) + x

    I'm not too confident about this solution however, if anyone could check and see if im right i'd appreciate it, and if I'm wrong highlight where you think ive made an error , thanks
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  2. #2
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    Differential equation

    Hello mitch_nufc
    Quote Originally Posted by mitch_nufc View Post
    find the general solution (x+y)y' = (3y-x), where y(1)=2

    I did the substitution y=ux

    I solved to get: ln|v-1| - 2/(v-1) = -ln|x| + A

    I then rearranged and introduced u=y/x back into the equation:

    ln|y-x| - 2x/(y-x) = A

    For y i get

    y = Ce^(2x/(y-x)) + x

    I'm not too confident about this solution however, if anyone could check and see if im right i'd appreciate it, and if I'm wrong highlight where you think ive made an error , thanks
    Yes, that looks OK to me. I differentiated your result, eliminated the term in Ce^{\frac{2x}{y-x}}, and got back to the original differential equation.

    Incidentally, you can use y(1) = 2 to get C = e^{-1}.

    Grandad
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  3. #3
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    Hello, mitch_nufc!

    Find the general solution: . (x+y)y' \:=\: (3y-x),\;\;\text{ where }y(1)=2

    I did the substitution:  y = vx

    I solved to get: . \ln|v-1| - \frac{2}{v-1} \:=\: -\ln|x| + A

    I then rearranged and introduced v=\tfrac{y}{x} back into the equation:
    . . \ln|y-x| - \frac{2x}{y-x}\:=\: A

    For y, i got: . y \:= \:Ce^{\frac{2x}{y-x}} + x . . . . So did I!

    Ya done good!

    Now use that initial condition to determine C.

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