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

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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large 14\color{royalblue}{\frac{dy}{dx}}=\color{orangered}{x+y}\] \[\large \color{orangered}{u=x+y}\]\[\large \frac{du}{dx}=1+\frac{dy}{dx} \qquad \rightarrow \qquad \color{royalblue}{\frac{dy}{dx}=\frac{du}{dx}1}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Understand how to plug in the substitution? :) I colorcoded it to make it a little easier.

dillpickles
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry 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/dx1=u?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes very good! :) We can use prime notation if it's a little easier to understand,\[\large 14(u'1)=u\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1From 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.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

dillpickles
 2 years ago
Best ResponseYou've already chosen the best response.0No, I understand that. I just have a hard time solving them on my own. Thanks

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Cool c: Lemme know if you get stuck.
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