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Thread: Please help solve ODE questions for DE newbie

  1. #1
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    Please help solve ODE questions for DE newbie

    1. y2(dx/dy) + xy -4y2 =1

    I try to simplify to

    (dx/dy) + x/y -4 = 1/y2

    (dy/dx) = y2 - x/y + 1/4

    I don't know how to go next (I try to use integrating factor but it's not working)

    2. dy/dx = y(y+3)

    (dy/dx)/y(y+3) = 1

    Integral (dy/dx)/y(y+3) = Integral 1 dx

    then I get

    1/3 In(y) - 1/3 In(y+3) = x + c
    (I'm not sure about this) My calculus is rusty. I don't remember how to get this number I just remember the formula
    Someone please explain or correct.

    Then I'm not sure how to get

    y = .......
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  2. #2
    Behold, the power of SARDINES!
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    Re: Please help solve ODE questions for DE newbie

    Quote Originally Posted by angelme View Post
    1. y2(dx/dy) + xy -4y2 =1

    I try to simplify to

    (dx/dy) + x/y -4 = 1/y2

    (dy/dx) = y2 - x/y + 1/4

    I don't know how to go next (I try to use integrating factor but it's not working)

    2. dy/dx = y(y+3)

    (dy/dx)/y(y+3) = 1

    Integral (dy/dx)/y(y+3) = Integral 1 dx

    then I get

    1/3 In(y) - 1/3 In(y+3) = x + c
    (I'm not sure about this) My calculus is rusty. I don't remember how to get this number I just remember the formula
    Someone please explain or correct.

    Then I'm not sure how to get

    y = .......
    For 1) the integrating factor should work.

    The integrating factor that I get is $\displaystyle \frac{1}{y}$ If you don't get this integrating factor post your work so we can see what went wrong.

    For 2)

    Multiply by 3 to get

    $\displaystyle \ln(y)+\ln(y+3)=3x+\hat{c}$

    $\displaystyle \ln(y(y+3))=3x+\hat{c} \iff y(y+3)=Ae^{3x}$

    This is quadratic in y. If you really want to isolate y you can use the quadratic formula.
    Thanks from angelme
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  3. #3
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    Re: Please help solve ODE questions for DE newbie

    For #1, let me use w instead of x (w=x). It hurts my eyes otherwise.

    $\displaystyle y^2w' + yw - 4y^2 = 1$.

    $\displaystyle w' + \frac{1}{y}w = 4 + y^{-2}$ (Doing this isn't ideal, since y=0 now becomes a bigger problem. But it's a minor concern.)

    Yes, the integrating factor is based on 1/y, so is actually $\displaystyle e^{\int \frac{dy}{y}} = e^{ln(|y|)+c} = A|y|$, so use $\displaystyle y$.

    With practice, that's "seeable" in the initial problem, $\displaystyle y^2w' + yw$. Here it amounts to observing that $\displaystyle yw' + w = (yw)'$.

    So only divide both sides of $\displaystyle y^2w' + yw = 1 + 4y^2$ by just one power of $\displaystyle y$ to get:

    $\displaystyle yw' + w = y^{-1} + 4y$, so $\displaystyle (yw)' = y^{-1} + 4y$. Then you're in business.

    ----------

    For #2, is it the partial fractions that you're unsure of? It works like this:

    Find $\displaystyle \int \frac{dy}{(y)(y+3)}$.

    Algebra: $\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3}$ for some constants $\displaystyle A$ and $\displaystyle B$ (it's the opposite of finding common demoninators.)

    It's then straightforward to find $\displaystyle A$ and $\displaystyle B$:

    $\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3}$ implies (clear demoninators)

    $\displaystyle 1 = A(y+3) + By$, implies $\displaystyle 1 = (A+B)y+3A$ for all y, implies $\displaystyle A+B = 0, 3A = 1$.

    Thus $\displaystyle A = \frac{1}{3}, B= \frac{-1}{3}$.

    Thus $\displaystyle \frac{1}{(y)(y+3)} = \frac{A}{y} + \frac{B}{y+3} = \frac{\frac{1}{3}}{y} + \frac{\frac{-1}{3}}{y+3}$

    Thus $\displaystyle \int \frac{dy}{(y)(y+3)} = \int \left( \frac{\frac{1}{3}}{y} + \frac{\frac{-1}{3}}{y+3} \right) dy$

    $\displaystyle = \frac{1}{3} \int \frac{1}{y} dy + \frac{-1}{3} \int \frac{1}{y+3} dy$

    $\displaystyle = \frac{1}{3} ln|y| - \frac{1}{3} ln|y+3| + C$

    (And you could simplify more: $\displaystyle = ln\left( \left| \frac{y}{y+3}\right|^{1/3} \right) + C$)
    Last edited by johnsomeone; Oct 5th 2012 at 09:58 PM.
    Thanks from angelme
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