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Thread: Ordinary Differential Equation (quick question about a step in the process)

  1. #1
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    Ordinary Differential Equation (quick question about a step in the process)

    Looking through my notes for a maths exam tomorrow I can't figure out the jump from one step to another:

    The original question is Solve:

    $\displaystyle \frac{dy}{dx}-y=x^2e^2x$

    now further down the steps I can't figure out how it jumps from:

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

    to

    $\displaystyle \frac{d}{dx}(y\frac{1}{x})=e^2x$

    What I want to know is how you get from the $\displaystyle \frac{dy}{dx}$ to the $\displaystyle \frac{d}{dx}$ and what happens to the $\displaystyle \frac{-1}{x^2}y$

    Thanks
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  2. #2
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    I think you have a typo in your problem, the original equation should be $\displaystyle xy'-y=x^2e^2x.$

    As for your question, what do you get if you try to differentiate w.r.t. to $\displaystyle x$ the expression $\displaystyle y\cdot\frac1x$ ?
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  3. #3
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    I don't see any typo no, that's the notation we use for Ordinary Differential Equations.

    Differentiating that will give you the $\displaystyle \frac{-1}{x^2}y$ yes, but what I think I'm meaning to say is, how do you jump from the $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{d}{dx}$ what is the difference between them.

    And if you do as you say and differentiate w.r.t. to the expression ? How can this still be the same as the line up above it? Surely that would make it: $\displaystyle \frac{-1}{x^2}y=e^2x$ and not

    Or am I completely missing something here?
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  4. #4
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    Quote Originally Posted by Macca567 View Post
    I don't see any typo no, that's the notation we use for Ordinary Differential Equations.

    Differentiating that will give you the $\displaystyle \frac{-1}{x^2}y$ yes, but what I think I'm meaning to say is, how do you jump from the $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{d}{dx}$ what is the difference between them.

    And if you do as you say and differentiate w.r.t. to the expression ? How can this still be the same as the line up above it? Surely that would make it: $\displaystyle \frac{-1}{x^2}y=e^2x$ and not

    Or am I completely missing something here?
    You are missing something.

    $\displaystyle
    \frac{d}{dx} \left(\frac{y}{x} \right) = \frac{x y' - y}{x^2}
    $

    by the quotient rule and implicit differentiation.
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