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Math Help - Differential Equation Questions

  1. #1
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    Differential Equation Questions

    I'm stuck on this problem

    Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

    first i move y and x around and get y(dy)=x(dx)

    then get the integral of both

    y^2/2 = x^2/2 + C

    multiply both sides by 2,

    y^2 = x^2 + C

    find C by plugging in 0 for X and -5 for Y

    -5^2 = 0 + C

    25 = C

    I plug C into the regular equation and

    then my final answer is y=sqrt( x^2 +25 )

    but it is wrong. What am i doing wrong? thanks!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by break View Post
    I'm stuck on this problem

    Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

    first i move y and x around and get y(dy)=x(dx)

    then get the integral of both

    y^2/2 = x^2/2 + C

    multiply both sides by 2,

    y^2 = x^2 + C

    find C by plugging in 0 for X and -5 for Y

    -5^2 = 0 + C

    25 = C

    I plug C into the regular equation and

    then my final answer is y=sqrt( x^2 +25 )

    but it is wrong. What am i doing wrong? thanks!
    Who says that's wrong? Looks A-okay to me
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  3. #3
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    Quote Originally Posted by AllanCuz View Post
    Who says that's wrong? Looks A-okay to me
    If y^2 = x^2 + 25 then surely y = \pm\sqrt{x^2 + 25}.
    Last edited by Prove It; May 28th 2010 at 09:28 PM.
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Prove It View Post
    If y^2 = x^2 - 25 then surely y = \pm\sqrt{x^2 - 25}.
    huh?

    y^2 = x^2 + C with the IC  y(0) = - 5

     25 = C

     y^2 = x^2 + 25

     y = \sqrt{ x^2 + 25}

    ?
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  5. #5
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    Quote Originally Posted by AllanCuz View Post
    huh?

    y^2 = x^2 + C with the IC  y(0) = - 5

     25 = C

     y^2 = x^2 + 25

     y = \sqrt{ x^2 + 25}

    ?
    Sorry, I put a minus where there should have been a plus. The point is, your work was fine, you just had to make y the subject.
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  6. #6
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    This is one of my online homework problems, apparently y=-sqrt(x^2+25) is correct. but i still don't understand why @_@.
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  7. #7
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    Quote Originally Posted by break View Post
    I'm stuck on this problem

    Find the solution of the differential equation that satisfies the given initial condition. (dy)/(dx) = x/y, y(0) = -5

    first i move y and x around and get y(dy)=x(dx)

    then get the integral of both

    y^2/2 = x^2/2 + C

    multiply both sides by 2,

    y^2 = x^2 + C

    find C by plugging in 0 for X and -5 for Y

    -5^2 = 0 + C

    25 = C

    I plug C into the regular equation and

    then my final answer is y=sqrt( x^2 +25 )

    but it is wrong. What am i doing wrong? thanks!
    The correct answer is y = {\color{red}-} \sqrt{x^2 + 25}, as is easily confirmed by substitution. Note that this solution satisfies y(0) = -5 ....
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