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Math Help - General Solution to Differential Equation

  1. #1
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    General Solution to Differential Equation

    Hi, I would be really grateful to anyone who may help with this...

    I need to find the general solution for the differential equation

    dy/dx = (cos x - sin x)e^cos x + sin x - y

    (giving the solution in implicent form)

    I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

    I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

    Many thanks for any help!
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  2. #2
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    Quote Originally Posted by looking0glass View Post
    Hi, I would be really grateful to anyone who may help with this...

    I need to find the general solution for the differential equation

    dy/dx = (cos x - sin x)e^cos x + sin x - y

    (giving the solution in implicent form)

    I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

    I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

    Many thanks for any help!
    Your typesetting is a little cumbetrsone and thus, hard to read. Is this your problem

     <br />
y' = \left(\cos x - \sin x\right)e^{\sin x + \cos x - y}\;?<br />
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  3. #3
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    Quote Originally Posted by looking0glass View Post
    Hi, I would be really grateful to anyone who may help with this...

    I need to find the general solution for the differential equation

    dy/dx = (cos x - sin x)e^cos x + sin x - y
    If this is \frac{dy}{dx}= (cos x- sin x)e^{cos x+ sin x- y}, as danny arigo suggests, then it is separable. \frac{dy}{dx}= (cos x- sin(x))e^{cos x+ sin x}e^{-y} so e^y dy= (cos x- sin x)e^{cos x + sin x} dx. Because, as you say, the derivative of e^{cos x+ sin x} is (cos x- sin x)e^{cos x+ sin x}, that's easy to integrate.

    (giving the solution in implicent form)

    I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

    I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

    Many thanks for any help!
    Integrating the equation in "separated" form above gives a "constant of integration". Use that fact that f(0)= 1 to determine that constant.
    Last edited by mr fantastic; May 31st 2009 at 03:47 AM. Reason: Added tag to close first quote.
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    Thanks for your help but I'm still not sure how to go about doing this...
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  5. #5
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    Quote Originally Posted by looking0glass View Post
    Thanks for your help but I'm still not sure how to go about doing this...
    As Hallsofivy said, the equation separates so

    <br />
e^y dy= (cos x- sin x)e^{cos x + sin x} dx

    so

    <br />
\int e^y dy= \int (cos x- sin x) e^{cos x + sin x} dx

    Let u = cos x + sin x so du = (cos x- sin x)dx so

    <br />
\int e^y dy= \int e^u du\;\;\; \Rightarrow\;\;\; e^y = e^u + c

    or

    <br />
e^y = e^{cos x + sin x} + c

    Now use the fact that y(0) = 1 to find c. Then take the \ln of both sides to find y.

    Hope that helps!
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