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Math Help - 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:

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

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

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

    to

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

    What I want to know is how you get from the \frac{dy}{dx} to the \frac{d}{dx} and what happens to the \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 xy'-y=x^2e^2x.

    As for your question, what do you get if you try to differentiate w.r.t. to x the expression 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 \frac{-1}{x^2}y yes, but what I think I'm meaning to say is, how do you jump from the \frac{dy}{dx} to \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: \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 \frac{-1}{x^2}y yes, but what I think I'm meaning to say is, how do you jump from the \frac{dy}{dx} to \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: \frac{-1}{x^2}y=e^2x and not

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

     <br />
\frac{d}{dx} \left(\frac{y}{x} \right) = \frac{x y' - y}{x^2}<br />

    by the quotient rule and implicit differentiation.
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