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Math Help - Separable Differential Equation problems

  1. #1
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    Separable Differential Equation problems

    I'm having trouble with the two following separable differential equation equation problems. I'm able to separate the variables, and even do the individual integration (I think the left side in both cases has to be integrated using partial fractions. If there is a different way, I would appreciate any guidance/suggestions). I'm just having trouble getting from the results of the initial integration to the final solution of the differential equations. Any help would be appreciated. Thank you very much in advance.

    1. Solve the given differential equation:





    2. Solve the given initial-value problem:

    ,
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  2. #2
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    Quote Originally Posted by rgriss1 View Post
    I'm having trouble with the two following separable differential equation equation problems. I'm able to separate the variables, and even do the individual integration (I think the left side in both cases has to be integrated using partial fractions. If there is a different way, I would appreciate any guidance/suggestions). I'm just having trouble getting from the results of the initial integration to the final solution of the differential equations. Any help would be appreciated. Thank you very much in advance.

    1. Solve the given differential equation:





    2. Solve the given initial-value problem:

    ,
    It's not clear exactly what your difficulty is. Yes, those are separable equations:
    The first is \frac{dy}{y^2- 1}= \frac{dx}{x^3}
    or \frac{dy}{(y-1)(y+1)}= x^{-3}dx
    Yes, the left side can be integrated using integration by parts.

    The second is \frac{dy}{y^2- 4}= dx
    or \frac{dy}{(y-2)(y+2)}= dx
    with condition y(0)= -6
    and the left side can be integrated using integration by parts.

    Perhaps you have having trouble solving for y as a function of x? That normally is not required for problems like these but can be done. For example, integrating the second problem, you get (1/4)ln(|y- 2|)- (1/4)ln(|y+2|)= x+ C. Setting y= -6 when x= 0 gives (1/4)ln(8)- (1/4)ln(4)= (1/4)(ln(8)- ln(4))= (1/4)ln(8/4)= (1/4)ln(2)= C so we have (1/4)ln(|y-2|)- (1/4)ln(|y+2|)= x+ (1/4)ln(2). which we can write as ln|(y-2)/(y+2)|= 4x+ ln 2. Taking the exponential of both sides, |(y-2)/(y+2)|= e^{4x+ ln 2}= 2 e^{4x}. Since the initial value is at y= -6 the solution can only be valid for y< -2 which makes both numerator and denominator negative, so the fraction is positive, so we can remove the absolute values and solve \frac{y-2}{y+2}= 2e^{4x} for y.
    Last edited by HallsofIvy; February 8th 2009 at 03:23 PM.
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