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Math Help - Sep. by Var.

  1. #1
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    Sep. by Var.

    I have another seperation by variables problem, and the hardest part for me is separating the variables which is probably why we have this lesson.

    This one seems complicated..I have to solve this dif eq by using sep. by variables:

    y dy = t(1 + t^2)^(-1/2) * (1 + y^2)^(1/2) dt

    The square root in the denominator for t and the square root in the numerator for y seems hard to separate them.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    I have another seperation by variables problem, and the hardest part for me is separating the variables which is probably why we have this lesson.

    This one seems complicated..I have to solve this dif eq by using sep. by variables:

    y dy = t(1 + t^2)^(-1/2) * (1 + y^2)^(1/2) dt

    The square root in the denominator for t and the square root in the numerator for y seems hard to separate them.
    y~dy = \frac{t}{\sqrt{1 + t^2}} \cdot \sqrt{1 + y^2}~dt

    \frac{y}{\sqrt{1 + y^2}}~dy = \frac{t}{\sqrt{1 + t^2}}~dt

    -Dan
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    Quote Originally Posted by topsquark View Post
    y~dy = \frac{t}{\sqrt{1 + t^2}} \cdot \sqrt{1 + y^2}~dt

    \frac{y}{\sqrt{1 + y^2}}~dy = \frac{t}{\sqrt{1 + t^2}}~dt

    -Dan
    Er duh! Thanks topsquark.

    One final question:

    Taking the integral of both sides yields:

    sqrt(y^2 + 1) = sqrt(t^2 + 1) + C. Now is THIS the solution or can we infer that y and t must be equal and say more?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Er duh! Thanks topsquark.

    One final question:

    Taking the integral of both sides yields:

    sqrt(y^2 + 1) = sqrt(t^2 + 1) + C. Now is THIS the solution or can we infer that y and t must be equal and say more?
    y = t only if C = 0
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    y = t only if C = 0
    Well, technically there is a "+ C" on both sides, but you can "group them up" and that's why my prof. says to just keep it on one side. So what if C was -5 on one side and 5 on the other. For simplicity, is my sol'n just sqrt(y^2 + 1) = sqrt(t^2 + 1) + C ?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Well, technically there is a "+ C" on both sides, but you can "group them up" and that's why my prof. says to just keep it on one side. So what if C was -5 on one side and 5 on the other. For simplicity, is my sol'n just sqrt(y^2 + 1) = sqrt(t^2 + 1) + C ?
    yes, but there is no guarantee that the C's on both sides are the same. i say play it safe. we cannot know for sure unless we were given some initial conditions
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