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Math Help - Separable Variables

  1. #1
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    Separable Variables

    I'm currently having trouble with these two problems:
    1. Solve the given differential equation by separation of variables
    (e^x + e^-x) dy/dx = y^2

    2. Find implicit and an explicit solution of the given IVP
    sqr root of (1-y^2) dx - sqr root of (1-x^2) dy = 0 ; y(0) = sqr root 3 / 2
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    Quote Originally Posted by element View Post
    I'm currently having trouble with these two problems:
    1. Solve the given differential equation by separation of variables
    (e^x + e^-x) dy/dx = y^2
    note that we can write this as

    \frac 1{y^2}~dy = \frac 1{e^x + e^{-x}}~dx

    now integrate both sides to continue

    2. Find implicit and an explicit solution of the given IVP
    sqr root of (1-y^2) dx - sqr root of (1-x^2) dy = 0 ; y(0) = sqr root 3 / 2
    note that we can write the equation as

    \frac 1{\sqrt{1 - y^2}}~dy = \frac 1{\sqrt{1 - x^2}}~dx

    now integrate both sides to continue
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    Quote Originally Posted by Jhevon View Post
    note that we can write this as

    \frac 1{y^2}~dy = \frac 1{e^x + e^{-x}}~dx

    now integrate both sides to continue

    note that we can write the equation as

    \frac 1{\sqrt{1 - y^2}}~dy = \frac 1{\sqrt{1 - x^2}}~dx

    now integrate both sides to continue
    I got that far for both 1 and 2. For 1, I'm stuck on integrating the right side. For 2, I got an answer but its not matching the books answer. Can you show me how you would do both? Thank you very much.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by element View Post
    I got that far for both 1 and 2. For 1, I'm stuck on integrating the right side.
    note that \int \frac 1{e^x + e^{-x}}~dx = \int \frac {e^x}{e^{2x} + 1}~dx .......(by multiplying by \frac {e^x}{e^x})

    now do a substitution of u = e^x


    For 2, I got an answer but its not matching the books answer. Can you show me how you would do both? Thank you very much.
    integrating both sides we get

    \arcsin y = \arcsin x + C

    \Rightarrow y = \sin (\arcsin x + C)

    now use your initial data to find the solution
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    Thank you very much. Just one quick general question, when finding the implicit and explicit solution, are we to always have "y" by itself?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by element View Post
    Thank you very much. Just one quick general question, when finding the implicit and explicit solution, are we to always have "y" by itself?
    "explicit" means y by itself, "implicit" means x's and y's mixed, or simply, not y by itself.
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