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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
    n/a


    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]
    \\ [/itex]

    Part B: First I find the general solution
    [tex]
    \begin{eqnarray}
    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 \\
    \end{eqnarray}
    [/tex]

    Then the particular solution
    [tex]
    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
    [/tex]
     
    Last edited: Apr 9, 2014
  2. jcsd
  3. Apr 9, 2014 #2

    vela

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    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

    vela

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    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 :)
     
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