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Math Help - Initial Value Problem

  1. #1
    Member VitaX's Avatar
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    Initial Value Problem

    Find the solution to the initial value problem in explicit form.

    y' = xy^3(1 + x^2)^{-\frac{1}{2}} ; y(0) = 1

    \int \frac{1}{y^3}dy = \int \frac{x}{(1+x^2)^{\frac{1}{2}}}dx

    -\frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2}} + C

    -2y^2 = \frac{1}{1 + x^2)^{\frac{1}{2}}} + C

    -y = \left(\frac{1}{2(1 + x^2)^{\frac{1}{2}}}\right)^{\frac{1}{2}} + C

    y = - \frac{1}{\left(2(1 + x^2)^{\frac{1}{2}}\right)^{\frac{1}{2}}} \right) + C

    I think I screwed up somewhere because when I go to solve for C I get C = 1 + \frac{1}{2^{\frac{1}{2}}}.

    The Book's answer in explicit form is y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}
    Last edited by VitaX; September 5th 2011 at 06:44 PM.
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  2. #2
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    Re: Initial Value Problem

    Quote Originally Posted by VitaX View Post
    Find the solution to the initial value problem in explicit form.

    y' = xy^3(1 + x^2)^{-\frac{1}{2}} ; y(0) = 1

    \int \frac{1}{y^3}dy = \int \frac{x}{(1+x^2)^{\frac{1}{2}}}dx

    -\frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2}} + C
    You could solve for C at this point.

    The next equation is incorrect.
    -2y^2 = \frac{1}{(1 + x^2)^{\frac{1}{2}}} + C
    It should be:
    -2y^2 = \frac{1}{(1 + x^2)^{\frac{1}{2}}+ C}

    -y = \left(\frac{1}{2(1+x^2)^{\frac{1}{2}}} \right)^{\frac{1}{2}} + C

    y = -\left(\frac{1}{2(1+x^2)^{\frac{1}{2}}} \right)^{\frac{1}{2}} + C

    I think I screwed up somewhere because when I go to solve for C I get C = 1 + \frac{1}{2^{\frac{1}{2}}}.

    The Book's answer in explicit form is y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}
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  3. #3
    Member VitaX's Avatar
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    Re: Initial Value Problem

    Why does C go in the denominator? I thought that since you are taking the reciprocal of both sides that \frac{1}{C} will just be another constant, C, so I left it alone. It's these types of things that are confusing me, the work is easy but knowing when C gets changed or not is a bit confusing here. Why exactly is C effected here if there is no variable effecting it?
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  4. #4
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    Re: Initial Value Problem

    It's pretty basic algebra !

    \frac{1}{a+b}\ne\frac{1}{a}+\frac{1}{b}
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  5. #5
    Member VitaX's Avatar
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    Re: Initial Value Problem

    Yeah I screwed up on that part.

    My main question is this though, after isolating y does it become this: y = \left [ \frac{-1}{2(1+x^2)^{\frac{1}{2}} + C}\right]^{\frac{1}{2}} or this: y = \left [ \frac{-1}{2(1+x^2)^{\frac{1}{2}} + 2C}\right]^{\frac{1}{2}}.

    I know this is hard to explain but I'll show a link to why I'm a bit confused with the subject of C.

    Problem 7 Page 4

    If you go to page 4 and problem 7, you can see where he raises both sides to e and e^C simply becomes C. So does 2C become C here or no? I only ask that because both e and 2 are constants, so isn't that the reason why in that problem it is written as C? The math is easy, but working with the C is confusing me with situations like this.
    Last edited by VitaX; September 5th 2011 at 09:27 PM.
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  6. #6
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    Re: Initial Value Problem

    He uses C (upper case) as the original constant and c (lower case) as the 'derived' constant. Therefore, he has c = e^C, where ^ denotes the exponentiation operation.

    In your case, you just need to be consistent about which way you use the constant, C, both when you plug-in the initial conditions to solve for C and then when you substitute that value of C back into your expression for y.

    You should get the same answer for y either way.
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