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Math Help - Show that the function y = f(x) is a solution of the given differential equation

  1. #1
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    Show that the function y = f(x) is a solution of the given differential equation

    y = \frac{1}{x} \int_1^x\frac {e^t}{t} dt ,  x^2y' + xy = e^x


    I tried to take the derivative of y equation to get the right equation but I'm keep getting (x^2)y' = \frac{-e^x}{x}
    Can someone explain this why I can't get to (x^2)y' + xy = e^x
    Last edited by fastcarslaugh; February 15th 2011 at 06:49 AM. Reason: to clarify the question
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  2. #2
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    Quote Originally Posted by fastcarslaugh View Post
    y = \frac{1}{x} \int_1^x\frac {e^t}{t} dt ,  x^2y' + xy = e^x


    I tried to take the derivative of y equation to get the right equation but I'm keep getting (x^2)y' = \frac{-e^x}{x}
    Can someone explain this why I can't get to (x^2)y' + xy = e^x
    You need to use the product rule when you take the derivative of

    \displaystyle y = \underbrace{\frac{1}{x}}_{f(x)} \underbrace{\int_1^x\frac {e^t}{t} dt}_{g(x)}

    So y'=f'(x)g(x)+f(x)g'(x)

    Now lets compute the parts

    \displaystyle f'(x)=-\frac{1}{x^2} and

    \displaystyle g'(x)=\frac{e^x}{x} This is from the Fundamental theorem of calculus.

    Can you finish from here?
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  3. #3
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    But how should I solve the \int_1^x\frac{e^t}{t}dt

    I am stuck at y' = \frac{-1}{x^2}\int_1^x\frac{e^t}{t}dt + \frac{e^x}{x^2}
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  4. #4
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    You don't need to evaluate that integral. It will "cancel" when you substitute into the DE.
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  5. #5
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    I tried this problem from another perspective.

    y = \frac{1}{x}\int_1^x\frac{e^t}{t}dt

    xy = \int_1^x\frac{e^t}{t}dt

    xy' + y = \frac{e^x}{x}

    x^2y' + xy = e^x

    This method that I used is a lot easier.



    Ps. should i make this solved?
    Also regarding TheEmptySet post, after looking at it third time that I realized that it just cancels out. So thank you TheEmptySet for your help.
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  6. #6
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    Quote Originally Posted by fastcarslaugh View Post
    I tried this problem from another perspective.

    y = \frac{1}{x}\int_1^x\frac{e^t}{t}dt

    xy = \int_1^x\frac{e^t}{t}dt

    xy' + y = \frac{e^x}{x}

    x^2y' + xy = e^x

    This method that I used is a lot easier.



    Ps. should i make this solved?
    Also regarding TheEmptySet post, after looking at it third time that I realized that it just cancels out. So thank you TheEmptySet for your help.
    Very elegant!
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