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Math Help - Ordinary Differential Equations

  1. #1
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    Ordinary Differential Equations

    Need help with following three differential equations. Help is really appreciated!

    First:
    Solve the differential equation  y' - 2 y  =  3, y(0) = 1.

    Second:

    \displaystyle \frac{dy}{dx} + 8 \sin(x) y = (x+3)^4 e^{8 \cos(x)}, y(0)=5.


    Third:

    \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2) that have the value 4 when x=0
    Last edited by mr fantastic; November 16th 2010 at 03:12 PM. Reason: Improved the latex and formatting.
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  2. #2
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    Quote Originally Posted by Anteeks View Post
    Need help with following three differential equations. Help is really appreciated!

    First:
    Solve the differential equation  y' - 2 y = 3, y(0) = 1.

    Second:

    \displaystyle \frac{dy}{dx} + 8 \sin(x) y = (x+3)^4 e^{8 \cos(x)}, y(0)=5.


    Third:

    \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2) that have the value 4 when x=0
    The first and third are seperable. The second uses the integrating factor.

    If you need more help, please show all your work and say where you're stuck.
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  3. #3
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    Thank you!

    First equation:

    e^int{(-2x)dx = e^{(-2x)

    \displaystyle \frac{dy}{dx}*e^{(-2x)) -2y*e^{(-2x) = 3*e^{(-2x)
    -->
    D(y*e^{(-2x)) =3*e^{(-2x)
    -->
    y*e^{(-2x)= int(3*e^{(-2x))dy =-6*e^{(-2x) + C

    y=-6+C*e^{(2x)
    y(0)=6 --> -6+C*e^0=6, C=11

    Which gives y=-6+11*e^{(2x)

    Is this right?

    Second:
    integrating factor =  e^{8 \cos(x)}

    \displaystyle \frac{dy}{dx} *e^{8 \cos(x)   + 8 \sin(x) * e^{8 \cos(x)  y = (x+3)^4 e^{8 \cos(x)}* e^{8 \cos(x)

    Thats about how far I can go, and don't know if its right.

    Third:
    \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)

    Divide with (3^2+y^2) on both rows and integrate both:

    \displaystyle \frac{dy}{dx}/(3^2+y^2) = x^5

    Thats how far I can go.
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  4. #4
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    Quote Originally Posted by Anteeks View Post
    Thank you!

    First equation:

    e^int{(-2x)}dx = e^{(-2x)
    Yes, that first equation is linear and so you can get an integrating factor this way. It is \int -2 dx= -2x, not \int -2x dx. After that you are good

    \displaystyle \frac{dy}{dx}*e^{(-2x)) -2y*e^{(-2x) = 3*e^{(-2x)
    -->
    D(y*e^{(-2x)) =3*e^{(-2x)
    -->
    y*e^{(-2x)= int(3*e^{(-2x))dy =-6*e^{(-2x) + C
    until here. 3\int e^{-2x}dx= -\frac{3}{2}e^{-2x}+ C, not -6 e^{-2x}.

    y=-6+C*e^{(2x)
    y(0)=6 --> -6+C*e^0=6, C=11
    But mr fantastic told you the equation was separable. It is much easier to do \frac{dy}{dx}= 2y+ 3 so that \frac{dy}{2y+ 3}= \frac{dx}. Integrating, ln(2y+ 3)= x+ C so that 2y+ 3= ce^{x}.

    Which gives y=-6+11*e^{(2x)
    Is this right?
    Well, it is not hard to check yourself: if y= -6+ 11e^{2x} then y'= -22e^{-2x} while -2y= 12- 22e^{-2x}. So y'- 2y= -22e^{-2x}+ 12- 22e^{-2x}= 12- 44e^{-2x}. No, that is not "3" and so your solution does not check.

    Second:
    integrating factor =  e^{8 \cos(x)}
    No, the integral of 8 sin(x) is -8 cos(x).

    \displaystyle \frac{dy}{dx} *e^{8 \cos(x)   + 8 \sin(x) * e^{8 \cos(x)  y = (x+3)^4 e^{8 \cos(x)}* e^{8 \cos(x)

    Thats about how far I can go, and don't know if its right.

    Third:
    \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)

    Divide with (3^2+y^2) on both rows and integrate both:

    \displaystyle \frac{dy}{dx}/(3^2+y^2) = x^5

    Thats how far I can go.
    One of the things you should have learned long ago is that \int \frac{du}{1^2+ u^2}= arctan(u)+ C.

    \frac{1}{3^2+ y^2}= \frac{1}{3^2}\frac{1}{1+ \left(\frac{y}{3}\right)^2}. Let u= y/3.
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