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Math Help - help with integration (differential equations)

  1. #1
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    help with integration (differential equations)

    solve the differential equation, subject to given condition

    (x^2)(y')+1/y^2=0 , y(1)=2

    any help would be appreciated. thanks!

    i ended up with -x^-2=y^3 and im stuck but im pretty sure im doing it wrong =/
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  2. #2
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    Hello, s0urgrapes!

    Solve the differential equation, subject to given condition
    . . x^2\!\cdot\!\frac{dy}{dx} + \frac{1}{y^2} \:=\:0,\quad y(1)\,=\,2

    Separate variables: . x^2\!\cdot\frac{dy}{dx} \:=\:-\frac{1}{y^2} \quad\Rightarrow\quad y^2dy \:=\:-x^{-2}dx

    Integrate: . \frac{1}{3}y^3 \;=\;x^{-1} + C\quad\Rightarrow\quad y^3 \:=\:\frac{3}{x} + C

    From y(1) = 2, we have: . 2^3 \:=\:\frac{3}{1} + C \quad\Rightarrow\quad C \,=\,5


    Therefore: . \boxed{y^3 \;=\;\frac{3}{x} + 5}

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  3. #3
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    wow thanks alot
    i think the mistake i made was making the -2 into a -3 while integrating instead of -1 like it should be
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