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Solve the differential equation 14y'= x + y
by making the change of variable u = x + y.
 one year ago
 one year ago
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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zepdrixBest ResponseYou've already chosen the best response.1
\[\large 14\color{royalblue}{\frac{dy}{dx}}=\color{orangered}{x+y}\] \[\large \color{orangered}{u=x+y}\]\[\large \frac{du}{dx}=1+\frac{dy}{dx} \qquad \rightarrow \qquad \color{royalblue}{\frac{dy}{dx}=\frac{du}{dx}1}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Understand how to plug in the substitution? :) I colorcoded it to make it a little easier.
 one year ago

dillpicklesBest ResponseYou've already chosen the best response.0
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/dx1=u?
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
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

zepdrixBest ResponseYou've already chosen the best response.1
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

dillpicklesBest ResponseYou've already chosen the best response.0
No, I understand that. I just have a hard time solving them on my own. Thanks
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Cool c: Lemme know if you get stuck.
 one year ago
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