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Math Help - Differential Equation

  1. #1
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    Post Differential Equation

    Find a particular solution from the differential equation to the following

    \frac{dy}{dx}  =  \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2

    I can't separate the variables, and also when I use y = vx the answers doesn't match the ans key which is y = \sqrt{5x^3 - x^2}

    Same problem with this question

    solve the DE : \frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}

    What method do we use for this type of DE?
    Last edited by Deci; February 6th 2014 at 03:59 AM.
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  2. #2
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    Re: Differential Equation

    Quote Originally Posted by Deci View Post
    Find a particular solution from the differential equation to the following

    \frac{dy}{dx}  =  \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2

    I can't separate the variables, and also when I use y = vx the answers doesn't match the ans key which is y = \sqrt{5x^3 - x^2}
    I get the answer key answer when I use y = xv. Post your work and maybe we can see where you are having trouble.

    Quote Originally Posted by Deci View Post
    Same problem with this question

    solve the DE : \frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}

    What method do we use for this type of DE?
    Same method should work. Let y = vx.
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  3. #3
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    Re: Differential Equation

    Quote Originally Posted by Deci View Post
    Find a particular solution from the differential equation to the following

    \frac{dy}{dx}  =  \frac{ x^2 + 3y^2 }{2xy} ; y(1) = 2

    I can't separate the variables, and also when I use y = vx the answers doesn't match the ans key which is y = \sqrt{5x^3 - x^2}

    Same problem with this question

    solve the DE : \frac{dy}{dx} = \frac{ x^2 - y^2 }{xy}

    What method do we use for this type of DE?
    See if you can write \displaystyle \begin{align*} \frac{dy}{dx} = f\left( \frac{y}{x} \right) \end{align*}. If so, then the substitution \displaystyle \begin{align*} u = \frac{y}{x} \end{align*} is appropriate.
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  4. #4
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    Re: Differential Equation

    1)
    \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}
    {2xy}\frac{dy}{dx} = {x^2 + 3y^2}

    y = vx
    \frac{dy}{dx} = v + x \frac{dv}{dx}

    2x (vx) (v+x \frac{dv}{dx}) = x^2 + 3 (vx)^2

    2v^2x^2 + 2vx^3 \frac{dv}{dx} = x^2 + 3v^2x^2

    2vx^3 \frac{dv}{dx} = x^2 + 3v^2x^2 - 2v^2x^2}

    2vx^3 \frac{dv}{dx} = x^2 (1 + v^2) (


     2vx^3 \frac{dv}{dx} = x^2(1 + v^2)

     x \frac{dv}{dx} = \frac{(1 + v^2)}{2v}

     \frac{1}{x} dx = \frac{2v}{1+ v^2} dv

     \frac{1}{x} dx = \frac{2v}{1+ v^2} dv

    integrate both sides

    ln|x| = ln | 1 + v^2|

    x = 1 + v^2

    v = \sqrt{1-x^2}

    y = x  \sqrt{1-x^2}
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  5. #5
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    Re: Differential Equation

    Quote Originally Posted by Deci View Post
    1)
    \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}
    {2xy}\frac{dy}{dx} = {x^2 + 3y^2}

    y = vx

    <snip>

    integrate both sides

    ln|x| = ln | 1 + v^2|

    x = 1 + v^2

    v = \sqrt{1-x^2}

    y = x  \sqrt{1-x^2}
    a few things

    you need to solve

    \left|x\right| = \left|1+v^2\right| so you'll have a second solution

    x=1+v^2 \Rightarrow v=\sqrt{x-1}$ not $v = \sqrt{1-x^2} as you've written.

    you've neglected the constant of integration. This needs to be solved for given the boundary condition.

    Start at the point where you integrate both sides and rework it.
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  6. #6
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    Re: Differential Equation

    Okay, I tried it again, please check whether my answer is correct or not

    laTexis not working properly, so I type it like this

    \frac{1}{x} dx = \frac{2v}{1+ v^2} dv

    integrateboth sides

    ln|x|= ln | 1 + v^2|

    x= 1 + v^2

    x= 1 + \frac({y}{x})^2 + C
    y(1)= 2 --> 1 = 1 + 4 + C
    C= -4
    x= 1 + v^2 - 4
    v= \sqrt{x+3}

    \frac{y}{x}= \sqrt{x+3}
    y= x \sqrt{x+3}
    y= \sqrt{x^3 + 3x^2}
    Last edited by Deci; February 8th 2014 at 05:29 AM.
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  7. #7
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    Re: Differential Equation

    No, that is not correct. The constant of integration comes when you integrate, not when you exponentiate. On line 4, you write "integrate both sides". The next line should read:

    ln|x| + C = ln|1+v^2|

    Note that if you add a constant to both sides, you can subtract it from the right and the constant on the left "absorbs" the one from the right.

    Now, when you exponentiate, you get:

    e^(ln|x| + C) = 1+v^2
    e^(ln|x|)e^C = 1+v^2
    xe^C = 1+v^2

    Now, I tend to write K for multiplicative constants. Let K = e^C:

    Kx = 1+v^2

    Now, Kx = 1 + (\frac{y}{x})^2, so y(1) = 2 --> K = 1+(\frac{2}{1})^2 = 5.

    5x = 1+v^2 implies v = \pm \sqrt{5x-1}

    y = xv = \pm x\sqrt{5x-1} = \pm \sqrt{5x^3-x^2}
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