1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equations - first order

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data

    a function has the feature that at any point, the product of its gradient and the x-cordinate is equal to the square root of the y-cordinate multiplied by 5.

    Part A: Write out a differential equation that describes the function
    Part B: If the curve passes through the point x=7/3, y=2, find the particular solution the differential equation.

    2. Relevant equations

    3. The attempt at a solution
    Right, I am extremely new to differential equations so go easy on me.

    What I have done is below, I would be very grateful if someone could check my workings please. Thanks :)

    p.s. sorry for the many steps, i like to do every little step

    Part A:
    \\ [/itex]

    Part B: First I find the general solution
    x (\frac{dy}{dx})&=&5\sqrt{y} \\
    x \,\, dy &=& 5\sqrt{y} \,\,dx \\
    dy&=&5\sqrt{y}x^{-1}\,\, dx \\
    \frac{1}{\sqrt{y}} \,\,dy &=& 5x^{-1} \,\,dx \\
    y^{-\frac{1}{2}} \,\,dy &=& 5x^{-1} \,\,dx \\
    \frac{1}{5}y^{-\frac{1}{2}} \,\, dy&=&x^{-1} \,\,dx \\
    \int\frac{1}{5}y^{-\frac{1}{2}} \,\, dy&=&\int x^{-1} \,\,dx \\
    \frac{1}{5}\int y^{-\frac{1}{2}} \,\, dy&=&\int x^{-1} \,\,dx \\
    \frac{1}{5}(2y^{\frac{1}{2}})&=&ln|x|+C \\
    \frac{2}{5}\sqrt{y}&=&ln|x|+C \\

    Then the particular solution
    C=\frac{2}{5}\sqrt{y}-ln|x| \\
    C=\frac{2}{5}\sqrt{2}-ln(\frac{7}{3}=-0.28 \\
    \therefore \frac{2}{5}\sqrt{y}=ln|x|-0.28
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'd write ## C = \frac 25 \sqrt 2 - \ln\frac 73##. Don't be so quick to plug things into a calculator.

    And you need to solve for ##y##. You should find
    $$y = \left(\frac 52 \ln \frac{x}{7/3} + \sqrt 2\right)^2.$$ You can see by inspection that ##y## satisfies the initial condition, and you can differentiate it to verify it satisfies the differential equation.
  4. Apr 10, 2014 #3
    Right ok, thanks. I didn't know I needed to solve for y.

    Could I right that logarithm as [itex]\ln \frac{3x}{7}[/itex] or is that not allowed because its in the logarithm ?

    Thanks :)
  5. Apr 10, 2014 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yeah, you can write the log that way. It just wrote it that way because it makes it really obvious that when x=7/3, the log vanishes, so y=2.
  6. Apr 10, 2014 #5
    Right I see, yeah that makes sense. Thanks :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Differential Equations order Date
Solving 2nd order DE with initial condition Jan 17, 2018
Solving 2nd order differential equation Dec 27, 2017
Wondering if these two First Linear Order IVPs are correct Sep 29, 2017