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Math Help - Initial value problem

  1. #1
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    Initial value problem

    Find the particular solution of the initial value problem.

    (a) dy/dx=e^-3x+2x; y(0)=5/3

    (b) given that y=sin(3x+4y) find dy/dx


    I have a test on monday and a topis similar to this will be included. I would really appreciate it if somebody could show me the workings and help me understand the math, so i can try some examples by myself.

    Thanks alott!!
    Last edited by jameslynskey; August 21st 2009 at 06:34 AM.
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  2. #2
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    Hi jameslynskey

    For (a) :
    \frac{dy}{dx}=e^{-3x}+2x

    dy=(e^{-3x}+2x)dx

    Integrate both sides. Information : y(0)=5/3 will be used to find the value of constant resulting from the integration.


    For (b) :
    You have to use implicit differentiation to find \frac{dy}{dx}. Do you know it ?


    I'm a little bit confused. Is the question asking to find the particular solution for question (b) ?

    Last edited by songoku; August 21st 2009 at 08:42 AM.
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  3. #3
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    could you please show me the working throught the integration method aswell please?

    And for part (b), i have worked through the 'Straud' mathematics book and figured it out (:

    Thanks
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  4. #4
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    Hi jameslynskey


    dy=(e^{-3x}+2x)dx

    \int dy=\int (e^{-3x}+2x)dx

    y=\int e^{-3x} \; dx \; + \int 2x \; dx

    I'm sure it's easy for you to find \int 2x \; dx

    As for \int e^{-3x} \; dx , can you find \int e^x\; dx ?
    Last edited by songoku; August 22nd 2009 at 08:15 AM.
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  5. #5
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    Thanks alot. I really appreachaite the help.

    One last question, what do i now have to do with the y(0)=5/3?

    Btw i am only a high school student i am just trying really hard so i can do well at A level.

    *Press Thanks button*
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  6. #6
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    Hi jameslynskey

    Please show me your integration result first because it will be easier to point out how to use y(0)=5/3

    P.S. I will press thanks button after you completely finish with this problem hahahaha
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  7. #7
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    Iv only been doing this for 2 days so beware lol...

    y=e^-3x+x^2

    and i meant i pressed thanks for you!
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  8. #8
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    Hi jameslynskey

    Your answer is partly right.

    Let :

    u = -3x

    \frac{du}{dx}= \; -3

    dx=\frac{du}{-3}

    So, \int e^{-3x} \; dx

    =\int e^u\; \frac{du}{-3}

    =-\frac{1}{3}\; \int e^u\; du

    Since \int e^x\; dx\; =\; e^x , then \int e^u\; du\; =\; e^u

    Hence, \int e^{-3x} \; dx\; =\; -\frac{1}{3}\; e^{-3x}


    The integration will be : y=-\frac{1}{3}\; e^{-3x}+x^2+c , where c is constant of integration

    Using information y(0) = 5/3, we can find the value of c. We substitute the value of x = 0 and y = 5/3 to find c. Can you find c ?

    P.S. I'm really going to press thanks button for you because of your effort
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  9. #9
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    y=-1/3e^-3x+x^2+c

    c=-1/3e^-3x+x^2-y ??

    c=-1/3e^-3(0)+(0)^2-5/3

    c=-2??
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  10. #10
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    Hi jameslynskey
    Quote Originally Posted by jameslynskey View Post

    c=-1/3e^-3x+x^2-y ??
    This is wrong. This should be : c = y+\frac{1}{3}e^{3x}-x^2

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  11. #11
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    c=y+1/3e^3x-x^2

    c=5/3+1/3e^3(0)-(0)^2

    c=2 ???
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  12. #12
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