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brinethery
Group Title
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
 one year ago
 one year ago
brinethery Group Title
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
 one year ago
 one year ago

This Question is Closed

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

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

Luigi0210 Group TitleBest ResponseYou've already chosen the best response.1
@Jhannybean
 one year ago

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

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

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

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

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

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

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
When 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?
 one year ago

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

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

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

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

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

brinethery Group TitleBest ResponseYou've already chosen the best response.1
Yes :). 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.
 one year ago

zepdrix Group TitleBest 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,\[\large \frac{du}{dt}=1+4u^2\]
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
I 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. :)
 one year ago

brinethery Group TitleBest ResponseYou've already chosen the best response.1
I'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.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
That 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
 one year ago

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

brinethery Group TitleBest ResponseYou've already chosen the best response.1
Zepdrix was right.
 one year ago

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

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

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

brinethery Group TitleBest ResponseYou've already chosen the best response.1
Thanks!
 one year ago
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