Results 1 to 12 of 12

Math Help - Help with finding general solution of differential equation

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    10

    Help with finding general solution of differential equation

    Hi

    I think I've managed to work out the first part of a question which is

    Use the Quotient Rule to differentiate the function h(x) = (1 + ln(x))/x (x>0)
    I get the answer ln(x)/x^2.

    Now the second part of the question is

    Using your answer to the first part, find the general solution of the differential equation

    dy/dx = -(ln(x)/x^2)y^1/2 (x > 0, y > 0) giving the answer in implicit form.


    Can you help?
    Last edited by flyingscotsman; May 29th 2008 at 11:11 AM. Reason: missed off sign from equation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2008
    Posts
    106
    The equation given can be solved using seperation of variables, you can write it as
    <br />
\int \frac{dy}{y^{\frac{1}{2}}}=\int -\frac{\ln x}{x^2}dx

    the integral on the LHS is trivial and can you see how to apply your answer to the first part to the RHS?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    10
    Thanks

    Ok, then dy/y^1/2 = -ln(x)/x^2 dx

    gives 1/(2/3)y^3/2 + a = -((1 + ln(x))/x) + b

    which is 1/(2/3)y^3/2 = -((1 + ln(x))/x) + c

    Is this right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by flyingscotsman View Post
    Thanks

    Ok, then dy/y^1/2 = -ln(x)/x^2 dx

    gives 1/(2/3)y^3/2 + a = -((1 + ln(x))/x) + b

    which is 1/(2/3)y^3/2 = -((1 + ln(x))/x) + c

    Is this right?
    Unfortunately not, no, but you're not too far off. You made two errors; one when finding the derivative of h(x), and another when trying to integrate the function of y.

    Quote Originally Posted by flyingscotsman View Post
    I think I've managed to work out the first part of a question which is

    Use the Quotient Rule to differentiate the function h(x) = (1 + ln(x))/x (x>0)
    I get the answer ln(x)/x^2.
    h(x)=\frac{1+\ln{x}}{x}

    \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

    u=1+\ln{x}

    \frac{du}{dx}=\frac{1}{x}

    v=x

    \frac{dv}{dx}=1

    \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}=\frac{x\frac{1}{x}-(1+\ln{x})}{x^2}

    This appears to be where you goofed:

    h'(x)=\frac{1-1-\ln{x}}{x^2}

    h'(x)=-\frac{\ln{x}}{x^2}

    Now the second part of the question is

    Using your answer to the first part, find the general solution of the differential equation

    dy/dx = -(ln(x)/x^2)y^1/2 (x > 0, y > 0) giving the answer in implicit form.
    \frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}

    -\frac{\ln{x}}{x^2}dx=\frac{1}{\sqrt{y}}dy

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

    h(x)=\int\frac{1}{\sqrt{y}}dy

    The above is fine, but to a beginner it may help to convert the square root to a fraction root:

    h(x)=\int y^{-\frac{1}{2}}dy

    Can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2008
    Posts
    10
    Hi

    Thanks for the correction in the first part.

    As for h(x) = y^-1/2 dy, is it 2y^1/2 after integration?

    Therefore giving as the particular solution

    2y^1/2 = -((1 + ln(x))/x) + c

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by flyingscotsman View Post
    Hi

    Thanks for the correction in the first part.

    As for h(x) = y^-1/2 dy, is it 2y^1/2 after integration?

    Therefore giving as the particular solution

    2y^1/2 = -((1 + ln(x))/x) + c

    Is this correct?
    Very close, but you're still forgetting your first error. When you find the derivative of h(x), you make a small mistake, which you need to fix.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2008
    Posts
    10
    Yes, I've made the mistake of including the minus.

    It should be 2y^1/2 = (1 + lnx)/x + c, without the minus.
    I hope that's it.

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by flyingscotsman View Post
    Yes, I've made the mistake of including the minus.

    It should be 2y^1/2 = (1 + lnx)/x + c, without the minus.
    I hope that's it.

    That is correct, yes.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2008
    Posts
    10

    Smile

    Many thanks for all your help.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    May 2008
    Posts
    19
    Quote Originally Posted by hatsoff View Post
    That is correct, yes.
    How did the lnX become 1+lnX again please?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by brainache View Post
    How did the lnX become 1+lnX again please?
    It was in the original equation:

    <br />
h(x)=\frac{1+\ln{x}}{x}<br />

    Which is the integral of part of the second equation:

    <br />
\frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}<br />
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    May 2008
    Posts
    19
    Quote Originally Posted by hatsoff View Post
    It was in the original equation:

    <br />
h(x)=\frac{1+\ln{x}}{x}<br />

    Which is the integral of part of the second equation:

    <br />
\frac{dy}{dx}=-\frac{\ln{x}}{x^2}\sqrt{y}<br />
    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 30th 2010, 05:30 PM
  2. finding general solution (difficult differential equation)
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: May 10th 2010, 11:22 PM
  3. general solution of a differential equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: December 5th 2009, 08:30 AM
  4. Finding a general solution to an exact differential equation.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 15th 2009, 03:39 PM
  5. General Solution of Differential Equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: November 17th 2008, 05:46 PM

Search Tags


/mathhelpforum @mathhelpforum