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Math Help - Finding general solution

  1. #1
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    Finding general solution

    Hello everyone!

    i was asked to use the quotient rule to differentiate
    h(x) = ( 1 + ln(x) ) / x where x > 0

    I got

    k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2

    k'(x) = ( ln(x) ) / x^2

    which i think is right.

    it then asks me to use the above answer to find the general soltion of the differentisl equation

    dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

    and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.
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  2. #2
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    Quote Originally Posted by Sweeties View Post
    Hello everyone!

    i was asked to use the quotient rule to differentiate
    h(x) = ( 1 + ln(x) ) / x where x > 0

    I got

    k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2

    k'(x) = ( ln(x) ) / x^2

    which i think is right.

    it then asks me to use the above answer to find the general soltion of the differentisl equation

    dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

    and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.
    You've found that the derivative of \frac{1 + \ln x}{x} is \frac{\ln x}{x^2}.

    Therefore \int \frac{\ln x}{x^2} \, dx = \frac{1 + \ln x}{x} + C.


    \frac{dy}{dx} = - \frac{\ln x}{x^2} \, y^{1/2} \Rightarrow - \frac{dy}{y^{1/2}} = \frac{\ln x}{x^2} \, dx \Rightarrow - \int \frac{dy}{y^{1/2}} = \int \frac{\ln x}{x^2} \, dx ......
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You've found that the derivative of \frac{1 + \ln x}{x} is \frac{\ln x}{x^2}.

    Therefore \int \frac{\ln x}{x^2} \, dx = \frac{1 + \ln x}{x} + C.


    \frac{dy}{dx} = - \frac{\ln x}{x^2} \, y^{1/2} \Rightarrow - \frac{dy}{y^{1/2}} = \frac{\ln x}{x^2} \, dx \Rightarrow - \int \frac{dy}{y^{1/2}} = \int \frac{\ln x}{x^2} \, dx ......
    Sory, its not clear to me why/ how you have got this:
    - dy/ y^(1/2)
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  4. #4
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    Quote Originally Posted by Sweeties View Post
    Sory, its not clear to me why/ how you have got this:
    - dy/ y^(1/2)
    Re-arrange. Put all the x-stuff on one side and all the y-stuff on the other. The "-" has to go somewhere. I put it with the y-stuff. If you prefer:

    \int \frac{dy}{y^{1/2}} = - \int \frac{\ln x}{x^2} \, dx

    is just as correct .....
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  5. #5
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    ok. i'm still not sure about the 'general solution in implicit form' part. thanks.
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  6. #6
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    Quote Originally Posted by Sweeties View Post
    ok. i'm still not sure about the 'general solution in implicit form' part. thanks.
    2 \sqrt{y} = \frac{1 + \ln x}{x} + C, perhaps?

    Although I don't think it's much of a stretch to get an explicit general solution for y .....
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  7. #7
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    uote=Sweeties;146108]Hello everyone!

    i was asked to use the quotient rule to differentiate
    h(x) = ( 1 + ln(x) ) / x where x > 0

    I got

    k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2 x + (1/x) this should be x*(1/x)
    according to the quotient rule
    k'(x) = ( ln(x) ) / x^2 this would then change the
    answer you got
    which i think is right.

    it then asks me to use the above answer to find the general soltion of the differentisl equation

    dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

    and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.[/quote]
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  8. #8
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    Quote Originally Posted by jo holvey View Post
    uote=Sweeties;146108]Hello everyone!

    i was asked to use the quotient rule to differentiate
    h(x) = ( 1 + ln(x) ) / x where x > 0

    I got

    k'(x) = ( x + (1/x) - 1 + ln(x) * 1 ) / x^2 x + (1/x) this should be x*(1/x)
    according to the quotient rule
    k'(x) = ( ln(x) ) / x^2 this would then change the
    answer you got
    which i think is right.

    it then asks me to use the above answer to find the general soltion of the differentisl equation

    dy/dx = -(( ln(x) ) / x^2)) y^1/2 where x>0 and y>0

    and give the solution in implicit form which i can't do. i am also not sure how to integrate ln(x), i find it confusing.
    I confess I didn't check the derivative properly. One of my many failings is a total trust in people when it comes to simple things

    In fact, the derivative of y = \frac{1 + \ln x}{x} is \frac{dy}{dx} = {\color{red}-}\frac{\ln x}{x^2}.

    The answer for the solution to the DE changes slightly as a consequence.
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