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Math Help - Separable Variables-Differential Equations

  1. #1
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    Separable Variables-Differential Equations

    Hello,

    I am having trouble solving the 2 following problems.

    Given the differential equation solve by separation of variables.

    1. sec^2(x) dy + csc(y) dx =0

    I have tried to get the y's on the left and the x's on the right and integrate, but I cannot get the correct solution. This is what I have so far....

    1/(csc(y)) dy = 1/(sec^2(x))

    But I'm just not getting 4cos(y)=2x+sin(2x)+c

    2. dy/dx = (xy + 3x - y-3) / (xy - 2x + 4y -8)

    On this problem I am not sure where to start.

    Any help is greatly appreciated. Thanks!
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    Quote Originally Posted by yellowrose View Post
    Hello,

    I am having trouble solving the 2 following problems.

    Given the differential equation solve by separation of variables.

    1. sec^2(x) dy + csc(y) dx =0

    I have tried to get the y's on the left and the x's on the right and integrate, but I cannot get the correct solution. This is what I have so far....

    1/(csc(y)) dy = 1/(sec^2(x))

    But I'm just not getting 4cos(y)=2x+sin(2x)+c
    \sec^2 x ~dy + \csc y ~dx = 0

    \Rightarrow \sec^2 x ~dy = - \csc y ~dx

    \Rightarrow \sin y ~dy = - \cos^2 x ~dx

    now continue
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  3. #3
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    Quote Originally Posted by yellowrose View Post
    Hello,

    I am having trouble solving the 2 following problems.

    Given the differential equation solve by separation of variables.

    1. sec^2(x) dy + csc(y) dx =0

    I have tried to get the y's on the left and the x's on the right and integrate, but I cannot get the correct solution. This is what I have so far....

    1/(csc(y)) dy = 1/(sec^2(x))

    Mr F says: \Rightarrow \sin y \, dy = \cos^2 x \, dx \Rightarrow \int \sin y \, dy = \int \cos^2 x \, dx.

    Note that from the double angle formula \cos (2x) = 2 \cos^2 x - 1, \cos^2 x = \frac{1}{2} (\cos(2x) + 1).

    This will all lead to
    [snip]4cos(y)=2x+sin(2x)+c
    [snip]
    ..
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    Quote Originally Posted by yellowrose View Post
    [snip]

    2. dy/dx = (xy + 3x - y-3) / (xy - 2x + 4y -8)

    On this problem I am not sure where to start.

    Any help is greatly appreciated. Thanks!
    Note that \frac{dy}{dx} = \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{y(x - 1) + 3(x - 1)}{y(x + 4) - 2(x + 4)}


    = \frac{y + 3}{y\left(\frac{x+4}{x-1} \right) - 2\left(\frac{x+4}{x-1} \right)} = \frac{y + 3}{\left(\frac{x+4}{x-1} \right) (y - 2)} = \left(\frac{x-1}{x+4} \right) \left( \frac{y + 3}{y - 2} \right).

    *Ahem* or you could just spot that \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{x(3 + y) - (3 + y)}{x(y - 2) + 4(y - 2)} = \frac{(x - 1)(3 + y)}{(x + 4)(y - 2)} .......

    I'll come back to this point if you still need help with it. Suffice to say, it's now separable .....
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    Quote Originally Posted by Jhevon View Post
    \sec^2 x ~dy + \csc y ~dx = 0

    \Rightarrow \sec^2 x ~dy = - \csc y ~dx

    \Rightarrow \sin y ~dy = - \cos^2 x ~dx

    now continue
    I integrated that and got
    -cos(y)=-.5(sin(x)cos(x)+x)+c

    I'm sorry I just can't figure out how to get this into simplier terms. So that I just have a y on the left.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowrose View Post
    I integrated that and got
    -cos(y)=-.5(sin(x)cos(x)+x)+c

    I'm sorry I just can't figure out how to get this into simplier terms. So that I just have a y on the left.
    well, you could take the arccosine of both sides (remove the minus signs). but i don't think you have to get the answer any simpler than what you have
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    Quote Originally Posted by mr fantastic View Post
    ..
    Thanks for your help. I understand that I need to use the rule you gave me, but I'm not sure how. Do I use this before or after I integrate? I guess I'm just really confused about this problem.
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    Quote Originally Posted by yellowrose View Post
    Thanks for your help. I understand that I need to use the rule you gave me, but I'm not sure how. Do I use this before or after I integrate? I guess I'm just really confused about this problem.
    The rule I gave for cos^2 x gets used before integrating .... it's used to facilitate the integration of cos^2 x .
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    Quote Originally Posted by mr fantastic View Post
    Note that \frac{dy}{dx} = \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{y(x - 1) + 3(x - 1)}{y(x + 4) - 2(x + 4)}


    = \frac{y + 3}{y\left(\frac{x+4}{x-1} \right) - 2\left(\frac{x+4}{x-1} \right)} = \frac{y + 3}{\left(\frac{x+4}{x-1} \right) (y - 2)} = \left(\frac{x-1}{x+4} \right) \left( \frac{y + 3}{y - 2} \right).

    *Ahem* or you could just spot that \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{x(3 + y) - (3 + y)}{x(y - 2) + 4(y - 2)} = \frac{(x - 1)(3 + y)}{(x + 4)(y - 2)} .......

    I'll come back to this point if you still need help with it. Suffice to say, it's now separable .....
    Some further help:


    \frac{dy}{dx} = \left(\frac{x-1}{x+4} \right) \left( \frac{y + 3}{y - 2} \right) \Rightarrow \int \frac{y - 2}{y + 3} \, dy = \int \frac{x-1}{x+4} \, dx.

    Note that:

    \frac{y - 2}{y + 3} = \frac{(y + 3) - 5}{y + 3} = 1 - \frac{5}{y + 3}


    \frac{x - 1}{x + 4} = \frac{(x + 4) - 5}{x + 4} = 1 - \frac{5}{x + 4}
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    Quote Originally Posted by mr fantastic View Post
    Note that \frac{dy}{dx} = \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{y(x - 1) + 3(x - 1)}{y(x + 4) - 2(x + 4)}


    = \frac{y + 3}{y\left(\frac{x+4}{x-1} \right) - 2\left(\frac{x+4}{x-1} \right)} = \frac{y + 3}{\left(\frac{x+4}{x-1} \right) (y - 2)} = \left(\frac{x-1}{x+4} \right) \left( \frac{y + 3}{y - 2} \right).

    *Ahem* or you could just spot that \frac{xy + 3x - y-3}{xy - 2x + 4y} = \frac{x(3 + y) - (3 + y)}{x(y - 2) + 4(y - 2)} = \frac{(x - 1)(3 + y)}{(x + 4)(y - 2)} .......

    I'll come back to this point if you still need help with it. Suffice to say, it's now separable .....
    I separated them, took the integral of both sides, raised it to the e, and got
    e^y(y+3)^5=e^x(x+4)^5
    but the ans states e^x(y+3)^5=e^y(x+4)^5
    could you tell me what I'm doing different....or is what I have still correct? Thanks!
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  11. #11
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    Quote Originally Posted by yellowrose View Post
    I separated them, took the integral of both sides, raised it to the e, and got
    e^y(y+3)^5=e^x(x+4)^5
    but the ans states e^x(y+3)^5=e^y(x+4)^5
    could you tell me what I'm doing different....or is what I have still correct? Mr F says: Your answer is not the same as the book's answer. The exponent terms are switched!
    Thanks!
    Quote Originally Posted by mr fantastic View Post
    Some further help:


    \frac{dy}{dx} = \left(\frac{x-1}{x+4} \right) \left( \frac{y + 3}{y - 2} \right) \Rightarrow \int \frac{y - 2}{y + 3} \, dy = \int \frac{x-1}{x+4} \, dx.

    Note that:

    \frac{y - 2}{y + 3} = \frac{(y + 3) - 5}{y + 3} = 1 - \frac{5}{y + 3}


    \frac{x - 1}{x + 4} = \frac{(x + 4) - 5}{x + 4} = 1 - \frac{5}{x + 4}
    So you have:

    \int 1 - \frac{5}{y + 3} \, dy = \int 1 - \frac{5}{x+4} \, dx


    \Rightarrow y - 5 \ln |y + 3| = x - 5 \ln |x + 4| + C


    \Rightarrow y + \ln |x + 4|^5 = x + \ln |y + 3|^5 + C


    Exponentiate both sides, and let A = e^C:


    \Rightarrow e^y (x + 4)^5 = Ae^x (y + 3)^5


    which is the same as you book's answer (except that your book did a very bad thing and either forgot to include the arbitrary constant of integration or conveniently set A = 1).

    I could guess what mistake you made, but I'll bet that the mistake is obvious in hindsight
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