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Thread: Differential Equations, what are they? a few ?'s

  1. #1
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    Differential Equations, what are they? a few ?'s

    I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

    first one:
    show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

    there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

    second:
    1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
    2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
    3) think of a solution NOT a meber of the family in part 2
    4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

    the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

    All help appreciated
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  2. #2
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    Quote Originally Posted by deltemis View Post
    I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

    first one:
    show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

    there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

    second:
    1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
    2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
    3) think of a solution NOT a meber of the family in part 2
    4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

    the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

    All help appreciated
    $\displaystyle x\,\frac{dy}{dx} + y = 2x$

    Can you see that the LHS is a product rule expansion of $\displaystyle \frac{d}{dx}\,(x\,y)$?


    So $\displaystyle \frac{d}{dx}\,(x\,y) = 2x$

    $\displaystyle x\,y = \int{2x\,dx}$

    $\displaystyle x\,y = x^2 + C$

    $\displaystyle y = x + Cx^{-1}$.


    If $\displaystyle C = -1$ then $\displaystyle x - x^{-1}$ is a solution.
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  3. #3
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    The derivative of $\displaystyle y = x - x^{-1}$ with respect to $\displaystyle x$ is $\displaystyle y' = 1 + \frac{1}{x^2}$.

    Substitute $\displaystyle y = x - x^{-1}$ and $\displaystyle y' = 1 + \frac{1}{x^2}$ in $\displaystyle xy' + y = 2x$. Simplify to an obviously true statement, such as $\displaystyle 0 = 0$.
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    wow thank you both for fast and simplified answers, I hadn't noticed that product expansion and I wouldn't have even thought of substitution

    I get it now, thank you both very much
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