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Math Help - Calculus Homework Help Please

  1. #1
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    Calculus Homework Help Please

    Find the general solution of the differential equation xy'+2y=0
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    Quote Originally Posted by vc15ao4 View Post
    Find the general solution of the differential equation xy'+2y=0
    If i'm not mistaken, i solved this problem already, anyway...
    we proceed using the separation of variables

    xy' + 2y = 0
    => y' + (2/x)y = 0
    => y' = (-2/x)y
    => y'/y = -2/x
    => lny = -2ln(x) + C = ln(x)^-2 + C.............integrated both sides
    since lny = lnAx^-2
    => y = Ax^-2 is the general solution, where A is a constant
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    Quote Originally Posted by vc15ao4 View Post
    Find the general solution of the differential equation xy'+2y=0
    Say x>0, otherwise it gets me angry

    Then,

    y'+(2/x)y=0

    y'=(-2/x)y

    Well y=0 is a solution. Assume y!=0.*

    y'/y=(-2/x)

    INT (y'/y) dx = INT (-2/x) dx

    ln |y| = -2*ln (x)+C

    y= exp(-2*ln(x)+C)=C/x^2 where C>0


    *)The case y=0 at a point does not need to be considered.
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  4. #4
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    i don't know

    can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer
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    Quote Originally Posted by vc15ao4 View Post
    Find the general solution of the differential equation xy'+2y=0
    This is separable:
    x(dy/dx) + 2y = 0

    x dy = -2y dx

    dy/y = -2 dx/x

    Int[dy/y] = -2*Int[dx/x]

    ln(y) = -2*ln(x) + C' <-- C' is a constant

    ln(y) + 2*ln(x) = C'

    ln(y*x^2) = C'

    y*x^2 = C <-- where C = e^{C'}

    y = C/x^2 where C is a constant that is not equal to 0.

    But note that y = 0 does, in fact, solve xy' + 2y = 0, so we can remove this restriction on C.

    -Dan
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    Quote Originally Posted by vc15ao4 View Post
    can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer
    That is because it is a solution.
    It is not general solution.

    Meaning, all solution.s
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by vc15ao4 View Post
    can somebody doublecheck their answer because my professor told me that y=x^-2 was not the correct answer
    the answer is a constant times x^-2 or C/x^2. you can check this answer yourself by finding the first derivative and plugging it in to the function
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  8. #8
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    thanks

    plugging in the first derivitive did help. what about this problem

    Evaluate the integral: int (t^2+1/t+1)dt
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    Quote Originally Posted by vc15ao4 View Post
    plugging in the first derivitive did help. what about this problem

    Evaluate the integral: int (t^2+1/t+1)dt
    (t^2+1)/(t+1)=(t^2+2t-2t/(t+1)=(t^2+2t)/(t+1)-2*(t)/(t+1)=t-2*(t/(t+1))

    But,
    t/(t+1)=(t+1-1/(t+1))=1-1/(t+1)

    Thus, the partial fractions are,

    t+1-1/(t+1)
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    Quote Originally Posted by vc15ao4 View Post
    plugging in the first derivitive did help. what about this problem

    Evaluate the integral: int (t^2+1/t+1)dt
    let u = t + 1
    => du = dt

    since u = t + 1
    => t = u - 1
    => t^2 = (u - 1)^2
    => t^2 + 1 = (u - 1)^2 + 1 = u^2 -2u + 2

    so our integral becomes:

    int{(u^2 -2u + 2)/u}du
    = int{u - 2 + 2/u}du
    = (1/2)u^2 - 2u + 2ln(u) + C
    = (1/2)(t + 1)^2 - 2(t + 1) + 2ln(t+1) + C
    = (1/2)(t + 1)( (t + 1) - 4) + 2ln(t + 1) + C
    = (1/2)(t + 1)(t - 3) + 2ln(t + 1) + C
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  11. #11
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    ???

    how come your answer and the perfecthacker's answer were so different
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    Quote Originally Posted by vc15ao4 View Post
    how come your answer and the perfecthacker's answer were so different
    that happens sometimes in doing integration, the difference will be absorbed in the arbitrary constants in our respective solutions (or possibly rectified through algebraic manipulations).

    my solution is actually the one given by The Integrator--Integrals from Mathematica
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    Quote Originally Posted by vc15ao4 View Post
    how come your answer and the perfecthacker's answer were so different
    i can come up with yet another solution if you want.

    was there a particular form you want the solution in?
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  14. #14
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    its ok

    no it should be fine but thanks anyway
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