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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

This is what I thought the answer is, but it's wrong. If it helps at all, I believe this is a Bernoulli equation. :-/
Proof that I actually tried hehe...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

What if we expanded the RHS first?
Try it... I am really unsure of what to do at this point.
Expand and use integrating factor
Ah...
Perhaps my math teacher led me astray. I'll give int factor a shot
I can't do it. I get stuck.
Lets see...Integrating factor is \(\large e^{\int p(t)y}\)\[\large \cfrac{dy}{dt} = t^2 +8yt+16y^2\]So you'll have \[\large \cfrac{dy}{dt} -8yt -16y^2 = t^2\]
But will we ever be able to isolate y with this method? That's what I wonder.
Yep.
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?
I personally don't like the "du/dt -1". But thats just my opinion.
So integrating factor would be e^(-4t^2) for this, and then just solve?
I think I am going to throw in the towel for tonight. Thank you all for your help anyway...
Try approaching the problem the way access suggested, it makes a lot more sense.
towel throwing time? :c aw
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.
\[\large u=t+4y\]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,\[\large \frac{du}{dt}=1+4u^2\]
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. :)
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.
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
That looks much better.
Zepdrix was right.
And accessdenied too :-). Thanks everyone!
You are welcome! Glad it was sorted out. :P
I like the way you worked ytour problem, @brinethery :)
Thanks!

Not the answer you are looking for?

Search for more explanations.

Ask your own question