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brinethery
 one year ago
Solve the separable differential equation for y by making the substitution u=t+4y
dy/dt = (t+4y)^2
Oh yeah, and y(0) = 7
brinethery
 one year ago
Solve the separable differential equation for y by making the substitution u=t+4y dy/dt = (t+4y)^2 Oh yeah, and y(0) = 7

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brinethery
 one year ago
Best ResponseYou've already chosen the best response.1This is what I thought the answer is, but it's wrong. If it helps at all, I believe this is a Bernoulli equation. :/

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1Proof that I actually tried hehe...

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0What if we expanded the RHS first?

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1Try it... I am really unsure of what to do at this point.

FutureMathProfessor
 one year ago
Best ResponseYou've already chosen the best response.0Expand and use integrating factor

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1Perhaps my math teacher led me astray. I'll give int factor a shot

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1I can't do it. I get stuck.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Lets see...Integrating factor is \(\large e^{\int p(t)y}\)\[\large \cfrac{dy}{dt} = t^2 +8yt+16y^2\]So you'll have \[\large \cfrac{dy}{dt} 8yt 16y^2 = t^2\]

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1But will we ever be able to isolate y with this method? That's what I wonder.

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1When you substitute: u=t+4y Could we not just take the derivative of both sides with respect to t: du/dt = 1 + 4 dy/dt and then solve for dy/dt in terms of du/dt to substitute as well? We then have: dy/dt = (t + 4y)^2 <==> 1/4(du/dt  1) = u^2 1/4 (du/dt  1) = u^2 du/dt  1 = 4u^2 du/dt = 4u^2 + 1 Because is that not in some sense a separable DE?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0I personally don't like the "du/dt 1". But thats just my opinion.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1So integrating factor would be e^(4t^2) for this, and then just solve?

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1I think I am going to throw in the towel for tonight. Thank you all for your help anyway...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Try approaching the problem the way access suggested, it makes a lot more sense.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2towel throwing time? :c aw

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1Yes :). I slept 2 hours last night (just couldn't fall asleep). My brain's not working right. Thank you all again, I always appreciate math help.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large u=t+4y\]Taking the derivative of both sides WRT y you get,\[\large \frac{du}{dy}=\frac{dt}{dy}+4\]"Dividing" through by dt/dy gives us,\[\large \frac{du}{dy}\cdot\frac{dy}{dt}=\color{orangered}{1}+4\frac{dy}{dt}\]OOoo Ok I think I see the mistake she made. Forgot the 1 up there, \[\large \frac{du}{dt}=\color{orangered}{1}+4\frac{dy}{dt}\]Which simplifies to,\[\large \frac{du}{dt}=1+4u^2\]

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1I think the d/dy (t) simply was taken to be 0, more or less like a partial derivative. Since the work only shows the du/dy = 4 Otherwise, it seems the work would have likely been clean. :)

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1I'll tell you what all: I'll ask my teacher in the morning how to approach this thing and I'll post the full solution tomorrow. Thank you all for your help.

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1That sounds like a fair deal. I am thinking this way looks most promising right now... \( \displaystyle \frac{du}{dt} = 1 + 4u^2 \) \( \displaystyle \frac{1}{1 + 4u^2} \; du = 1 \; dt \) \( \displaystyle \int \frac{1}{1 + 4u^2} \; du = \int 1 \; dt \) \( \displaystyle \frac{1}{2} \tan^{1} \left(2u\right) = t + C \) \(u = t + 4y \) \( \displaystyle \frac{1}{2} \tan^{1} \left( 2(t + 4y)\right) = t + C \) I think that looks fairly reasonable. :P

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0That looks much better.

brinethery
 one year ago
Best ResponseYou've already chosen the best response.1And accessdenied too :). Thanks everyone!

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1You are welcome! Glad it was sorted out. :P

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0I like the way you worked ytour problem, @brinethery :)
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