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Math Help - Separable Equation ..

  1. #1
    Ted
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    Separable Equation ..

    Hello.

    Problem: Find the solution of the differenial equation that satisfies the given intial condition.
    \left( 2y + e^{3y} \right) \frac{dy}{dx} = x cos(x)
    The intial condition: y(0)=0.

    My Solution:

    I rewrite it as:

    \left( 2y + e^{3y} \right)dy=xcos(x)dx

    By integrating both sides and using integration by parts for the right side, I obtained:

    y^2+\frac{1}{3}e^{3y}=xsin(x)+cos(x)+C.

    The problem here is that I can not solve the last equation for y so that I can use the intial condition to find the particular solution.
    Any help?
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  2. #2
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    Quote Originally Posted by Ted View Post
    The problem here is that I can not solve the last equation for y so that I can use the intial condition to find the particular solution.
    Any help?
    You can use the initial condition at this point to find c. You just won't be able to put it into the form y = f(x)
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  3. #3
    Ted
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    Quote Originally Posted by pickslides View Post
    You can use the initial condition at this point to find c. You just won't be able to put it into the form y = f(x)
    Hmmm.
    OK I will put x=0.
    What about the y's ?
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  4. #4
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    Quote Originally Posted by Ted View Post
    The intial condition: y(0)=0.
    This means when x=0 then y = 0
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  5. #5
    Ted
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    Quote Originally Posted by pickslides View Post
    This means when x=0 then y = 0
    Sorry I did not get this one.
    What I know is: y(0)=0 solve the equation for y, then in the resulting function put x=0 and equate it to zero, and solve the resulting equation for the constant and substitute the value of the constant in the general solution to get the particular solution.
    I did not understand why this means I should put y=0.

    please, anyone can explain it?
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  6. #6
    o_O
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    y is assumed to be a function of x, i.e. {\color{red}y} = y({\color{blue}x}).

    The initial condition says that y({\color{blue}0}) = {\color{red}0}. This means that your particular solution has to pass through the point ({\color{blue}x},{\color{red}y}) = ({\color{blue}0},{\color{red}0}) .

    You don't have to solve for y explicitly. All you have to do is plug in x=0 and y=0 and solve for C.

    So: ({\color{red}0})^2+\frac{1}{3}e^{3({\color{red}0})  } = {\color{blue}0}\sin ({\color{blue}0}) +  \cdots etc.
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