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Thread: Differential equations

  1. #1
    Sep 2007

    Differential equations

    I'm after some help with these two equations.

    My problem is factorising with equations like this, I can never tell when I'm supposed to do it!! I have the answers to these questions I just need explicit help, basically a walk through of these examples, if anyone can help I'd much appreciate it,


    which satisfies y(0)=4

    The Answer is:
    sqrt(16-(14/3x^3 + 7x^2+10x))

    and my other problem is,

    dy/dx = 1+y^2/7+9x

    which satisfies y(1) = 0


    With regards to the last answer, how does PI sometimes come out of this answer?

    PS how do you write with math symbols, it looks a bit messy....
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Mar 2007
    Santiago, Chile
    Both of them are separable.

    (See my signature for LaTeX typesetting.)
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  3. #3
    Super Member wingless's Avatar
    Dec 2007
    A separable differential equation is an equation that can be expressed in this form:
    $\displaystyle \frac{dy}{dx} = g(x)\cdot h(y)$

    Like, $\displaystyle \frac{dy}{dx} = 4xy^2,~ \frac{dy}{dx} = \frac{x^2y-y}{x+1},~ \frac{dy}{dx} = \frac{xy^3}{\sqrt{x^2+1}}~.~.~.$

    You can solve them by separating the functions and differentials to the sides and integrating.

    For example,

    $\displaystyle \frac{dy}{dx} = 2xy^2$

    Move $\displaystyle dx$ to the right and $\displaystyle y^2$ to the left.

    $\displaystyle \frac{1}{y^2}~dy = 2x~dx$

    Now you can integrate both sides.

    $\displaystyle \int \frac{1}{y^2}~dy = \int 2x~dx$

    $\displaystyle -\frac{1}{y} + C_1 = x^2 + C_2$

    We can use only C instead of two integration constants.
    $\displaystyle -\frac{1}{y} = x^2 + C$

    $\displaystyle y = - \frac{1}{x^2 + C}$


    Your first question is,
    $\displaystyle y\frac{dy}{dx} +7x^2+7x+5=0$, $\displaystyle y(0)=4$

    Start by separating.

    $\displaystyle y~dy = -7x^2-7x-5~dx$

    Now we can integrate both sides.

    $\displaystyle \int y~dy = \int -7x^2-7x-5~dx$

    $\displaystyle \frac{y^2}{2}+C_1 = -\frac{7}{3}x^3 -\frac{7}{2}x^2-5x +C_2$

    $\displaystyle \frac{y^2}{2} = -\frac{7}{3}x^3 -\frac{7}{2}x^2-5x +C$

    $\displaystyle y^2 = -\frac{14}{3}x^3 -7x^2-10x +2C$

    $\displaystyle |y| = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

    -----$\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

    -----$\displaystyle y = -\sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

    Try $\displaystyle y(0) = 4$ for both of them.

    -----$\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$
    -----$\displaystyle 4 = \sqrt{2C}$
    -----$\displaystyle \boxed{2C = 16}$

    -----$\displaystyle y = -\sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$
    -----$\displaystyle 4 = -\sqrt{2C}$
    -----As $\displaystyle \sqrt{2C}$ is always positive, this equation doesn't have a root.

    So we found that our function is $\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$ because this is the only valid function that makes $\displaystyle y(0) = 4$. And we found that $\displaystyle 2C = 16$.

    Hence, the function we are looking for is:
    $\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x + 16}$

    $\displaystyle y = \sqrt{16-(\frac{14}{3}x^3 +7x^2+10x)}$


    Now try to solve the second equation
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