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dillpickles Group Title

Solve the differential equation 14y'= x + y by making the change of variable u = x + y.

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    \[\large 14\color{royalblue}{\frac{dy}{dx}}=\color{orangered}{x+y}\] \[\large \color{orangered}{u=x+y}\]\[\large \frac{du}{dx}=1+\frac{dy}{dx} \qquad \rightarrow \qquad \color{royalblue}{\frac{dy}{dx}=\frac{du}{dx}-1}\]

    • one year ago
  2. zepdrix Group Title
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    Understand how to plug in the substitution? :) I color-coded it to make it a little easier.

    • one year ago
  3. dillpickles Group Title
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    Sorry but I am still confused. I understand where everything came from but what do i do from where you left off? Do I write 14du/dx-1=u?

    • one year ago
  4. zepdrix Group Title
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    Yes very good! :) We can use prime notation if it's a little easier to understand,\[\large 14(u'-1)=u\]

    • one year ago
  5. zepdrix Group Title
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    From here, distribute the 14 to each term in the brackets. Then you can solve it either by separation of variables, or by getting it into standard form and finding an integrating factor.

    • one year ago
  6. zepdrix Group Title
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    I hope this step wasn't too confusing, \(\large \frac{du}{dx}=1+\frac{dy}{dx}\) It's the same as \(\large u'=1+y'\) I was just trying to emphasize that we're taking the derivative with respect to x. That's what the Leibniz notation was showing us.

    • one year ago
  7. dillpickles Group Title
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    No, I understand that. I just have a hard time solving them on my own. Thanks

    • one year ago
  8. zepdrix Group Title
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    Cool c: Lemme know if you get stuck.

    • one year ago
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