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Thread: Differential Equation

  1. #1
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    Differential Equation

    Hello everyone!

    Can somebody help me out with this one please...
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  2. #2
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    It's a "Show That". All you need is a couple of derivatives and some algebra.

    Let's see what you get for the 1st and 2nd derivative of the provided expression.
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  3. #3
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    for part a) Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ and substitute it into the equation given you should find that the LHS=RHS which shows its a solution.

    for the second part you need to solve the complimentary function
    $\displaystyle \frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$

    this would have an auxilary quadratic of
    $\displaystyle \lambda^2-2\lambda+1=0$

    solving that you get $\displaystyle \lambda=1$ as a repeated root so the solution to the complimentary function is

    $\displaystyle y=e^x(A+Bx)$ where A and B are the constants of integration.

    Putting this together with the solution we were given in part a) we have

    $\displaystyle y=e^x(A+Bx)+\frac{1}{2}x^2e^x$

    so all we have to do is find the constants from the given boundary conditions.

    I get A=1 and B=1 do post again if you have any problem with doing that or understanding what I've written

    Hope that helps

    Simon
    Last edited by thelostchild; May 28th 2008 at 09:36 AM.
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  4. #4
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    I couldnt get the ansa as e^x
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  5. #5
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    your differentiating wrong do you know the product rule?

    When u and v are functions of x
    $\displaystyle
    \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

    in this case we would have $\displaystyle u=\frac{1}{2}x^2$ and $\displaystyle v=e^x$

    so using the product rule

    $\displaystyle \frac{dy}{dx}=\frac{1}{2}x^2e^x+xe^x$

    can you do it the second time to get $\displaystyle \frac{d^2y}{dx^2}$ now?
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  6. #6
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    yeh, I get really confused knowing when to use which rule I can see why its the product rule here but would I now use the chain rule to diffeentiate again because the result isnt a product now
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  7. #7
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    no you would use the product rule but twice remember that

    $\displaystyle \frac{d}{dx}(\frac{1}{2}x^2e^x+xe^x)=\frac{d}{dx}( \frac{1}{2}x^2e^x)+\frac{d}{dx}(xe^x)$
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  8. #8
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    I think i went wrong sumplace...
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  9. #9
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    Quote Originally Posted by Sweeties View Post
    I think i went wrong sumplace...
    Hello,

    In the last line :

    $\displaystyle xe^x+xe^x+e^x=(2x+1)e^x \neq 3xe^x$



    Also, be careful when using tex for multiplications, don't use x because it's just like the variable and we can easily get confused. You can use $\displaystyle \cdot$ ( \cdot) or $\displaystyle \times$ ( \times)

    Plus, you made some typos like d/dx (uv)=(xe^x), which is false to write. I guess you wanted to say :

    $\displaystyle \frac{d}{dx} (xe^x)=\frac{d}{dx} (uv)$

    Want me to try showing you how to write all this stuff, or is it ok ?
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  10. #10
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    Thanks Moo
    I am using word 2007 with microsoft math add-in, just getting used to it
    i havent got latex on this computer, been meaning to download it!

    I think I got it now.
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  11. #11
    Moo
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    Quote Originally Posted by Sweeties View Post
    Thanks Moo
    I am using word 2007 with microsoft math add-in, just getting used to it
    i havent got latex on this computer, been meaning to download it!

    I think I got it now.
    You just have to use [tex][/tex] here

    And know some things about the latex code (there are topics in the latex help forum )
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  12. #12
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    Quote Originally Posted by Moo View Post
    You just have to use [tex][/tex] here

    And know some things about the latex code (there are topics in the latex help forum )
    Wicked!!! I just went through the tutorial. Thanks moo!!!
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