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Math Help - [SOLVED] Particular solution of a diffy eq

  1. #1
    Moo
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    [SOLVED] Particular solution of a diffy eq

    Hi

    There's this differential equation :
    \cosh(x) y'-\sinh(x) y=\sinh(x)
    Where y'=\frac{dy}{dx}

    The general solution is y=k \cosh(x)+y_0
    where y_0 is a defined function, solution of the equation.
    But I can't find it
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    Quote Originally Posted by Moo View Post
    Hi

    There's this differential equation :
    \cosh(x) y'-\sinh(x) y=\sinh(x)
    Where y'=\frac{dy}{dx}

    The general solution is y=k \cosh(x)+y_0
    where y_0 is a defined function, solution of the equation.
    But I can't find it
    Divide everything through by \cosh{x} and then it is first order linear, which you can solve using the integrating factor method.

    Note: \int{\tanh{x}\,dx} = \log{\cosh{x}}.
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    Moo
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    Quote Originally Posted by Prove It View Post
    Divide everything through by \cosh{x} and then it is first order linear, which you can solve using the integrating factor method.

    Note: \int{\tanh{x}\,dx} = \log{\cosh{x}}.
    y'=\tanh(x) y+\sinh(x)

    But I don't see how to solve it.


    Actually, this is how I found the general solution :
    we know that if we substitute by z=y-y_0, where y_0 is a particular solution, we get :
    \cosh(x) z'-\sinh(x) z=0
    This clearly gives z=k \cosh(x) (using this integrating factor)

    So the general solution is y=k \cosh(x)+y_0.

    But the problem is to find y_0 ><
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    Quote Originally Posted by Moo View Post
    y'=\tanh(x) y+\sinh(x)

    But I don't see how to solve it.


    Actually, this is how I found the general solution :
    we know that if we substitute by z=y-y_0, where y_0 is a particular solution, we get :
    \cosh(x) z'-\sinh(x) z=0
    This clearly gives z=k \cosh(x) (using this integrating factor)

    So the general solution is y=k \cosh(x)+y_0.

    But the problem is to find y_0 ><
    \frac{dy}{dx} -\tanh{(x)}y = \tanh{x}.

    The integrating factor is e^{\int{-\tanh{x}\,dx}} = \frac{1}{\cosh{x}} (Check).

    So using the integrating factor method

    \frac{d}{dx}\left(\frac{y}{\cosh{x}}\right) = \frac{\sinh{x}}{\cosh^2{x}}.

    Can you go from here?
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    Moo
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    Quote Originally Posted by Prove It View Post
    \frac{dy}{dx} -\tanh{(x)}y = \tanh{x}.

    The integrating factor is e^{\int{-\tanh{x}\,dx}} = \frac{1}{\cosh{x}} (Check).

    So using the integrating factor method

    \frac{d}{dx}\left(\frac{y}{\cosh{x}}\right) = \frac{\sinh{x}}{\cosh^2{x}}.

    Can you go from here?
    if we solve for y in the latter expression, we should get y=-1 ? oO

    What a fool... Thanks ^^

    but though, can you explain how to get the latter expression ?
    do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)
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    Quote Originally Posted by Moo View Post
    if we solve for y in the latter expression, we should get y=-1 ? oO

    What a fool... Thanks ^^

    but though, can you explain how to get the latter expression ?
    do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)
    Yes I get y = -1 + C.

    Integrating factor - Wikipedia, the free encyclopedia

    Oops, didn't see your part about wikipedia.

    Hang on...
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    Quote Originally Posted by Prove It View Post
    Yes I get y = -1 + C.

    Integrating factor - Wikipedia, the free encyclopedia

    Oops, didn't see your part about wikipedia.

    Hang on...
    Try here

    http://www.rdg.ac.uk/AcaDepts/sp/PPL..._integfact.pdf
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    Moo
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    Quote Originally Posted by Prove It View Post
    Thanks
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    Quote Originally Posted by Moo View Post
    Hi

    There's this differential equation :
    \cosh(x) y'-\sinh(x) y=\sinh(x)


    Let y=f(x) be a solution.
    Then it means, \cosh (x) f ' (x) - \sinh (x) f( x) = \sinh (x) for all x\in \mathbb{R}.
    Thus, \cosh (-x) f ' (-x) - \sinh (-x) f( -x) = \sinh (-x) for all x\in \mathbb{R}.
    Thus, \cosh (x) f ' (-x) + \sinh (x) f( - x) = - \sinh (x) for all x\in \mathbb{R}.

    What have we established, Moo?
    We have shown that if f(x) solves \cosh (x) y' - \sinh (x) y = \sinh (x)
    Then f(-x) solves \cosh (x) y' + \sinh (x) y = -\sinh (x).

    Now, \cosh (x) y' + \sinh (x) y = - \sinh (x) \implies [ \sinh (x) y] ' = - \sinh (x) [1]

    Can you take it from there?
    (When you solve [1] it would mean y(-x) solves the original problem).
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    Quote Originally Posted by Moo View Post
    do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)
    Ummmm...my tutorial!!!!
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