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Math Help - Help with one simple problem

  1. #1
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    Help with one simple problem

    Help with one simple problem-img010.jpg
    Here I am supposed to show that the equation of the right is a solution of the differential equation on the left. I have solved a few problems of this type already but I have trouble with this one. Did I make a mistake and overlooked something somewhere or the condition 'for x>0' sort of means that it is okay even if the other side of the equation turns out to be negative? In the book's answer key it says that for this problem the equation on the right is a solution of the DE
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  2. #2
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    Re: Help with one simple problem

    There must be a typo in here somewhere. I'm showing the solution to this differential equation is what mathematica calls

    $C x+\frac{1}{2} x \left(Ei(x)-\frac{e^x}{x}\right)$

    $Ei(x)=\int_{-x}^\infty \dfrac {e^{-t}} t~dt$
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    Re: Help with one simple problem

    Quote Originally Posted by romsek View Post
    There must be a typo in here somewhere.
    Oh so you mean that perhaps there was a typo in the book.. I have after all been able to solve the other problems..
    Can you show me where you got that solution by the way so I may make use of it too?
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    Re: Help with one simple problem

    Quote Originally Posted by catenary View Post
    Click image for larger version. 

Name:	img010.jpg 
Views:	8 
Size:	433.5 KB 
ID:	31153
    Here I am supposed to show that the equation of the right is a solution of the differential equation on the left. I have solved a few problems of this type already but I have trouble with this one. Did I make a mistake and overlooked something somewhere or the condition 'for x>0' sort of means that it is okay even if the other side of the equation turns out to be negative? In the book's answer key it says that for this problem the equation on the right is a solution of the DE
    You may have missed a negative sign... If the original problem were y' = -\dfrac{2y+e^x}{2x}, then y = \dfrac{C-e^x}{2x}. Are you sure you didn't miss a negative sign in the original problem?
    Last edited by SlipEternal; June 22nd 2014 at 01:10 AM.
    Thanks from catenary
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    Re: Help with one simple problem

    Oh it seems that it was not my solution that was wrong but rather I forgot to copy the negative sign in y' = -\dfrac{2y+e^x}{2x} when I copied it on the paper.. Thanks again so much SlipEternal you are my savior! By the way what is the difference between\dfrac and \frac?
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  6. #6
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    Re: Help with one simple problem

    \dfrac is a fraction in displaystyle - so usually appears bigger. But if you use the \displaystyle command at the beginning of your LaTeX that makes everything large size anyway
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