Question

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

Answer 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. :-/

Answer 2

Proof that I actually tried hehe...

Answer 3

@Jhannybean

Answer 4

What if we expanded the RHS first?

Answer 5

Try it... I am really unsure of what to do at this point.

Answer 6

Expand and use integrating factor

Answer 7

Ah...

Answer 8

Perhaps my math teacher led me astray. I'll give int factor a shot

Answer 9

I can't do it. I get stuck.

Answer 10

Lets see...Integrating factor is \(\large e^{\int p(t)y}\)\[\large \cfrac{dy}{dt} = t^2 +8yt+16y^2\]So you'll have \[\large \cfrac{dy}{dt} -8yt -16y^2 = t^2\]

Answer 11

But will we ever be able to isolate y with this method? That's what I wonder.

Answer 12

Yep.

Answer 13

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?

Answer 14

I personally don't like the "du/dt -1". But thats just my opinion.

Answer 15

So integrating factor would be e^(-4t^2) for this, and then just solve?

Answer 16

I think I am going to throw in the towel for tonight. Thank you all for your help anyway...

Answer 17

Try approaching the problem the way access suggested, it makes a lot more sense.

Answer 18

towel throwing time? :c aw

Answer 19

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.

Answer 20

\[\large u=t+4y\]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,\[\large \frac{du}{dt}=1+4u^2\]

Answer 21

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. :)

Answer 22

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.

Answer 23

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

Answer 24

That looks much better.

Answer 25

Zepdrix was right.

Answer 26

And accessdenied too :-). Thanks everyone!

Answer 27

You are welcome! Glad it was sorted out. :P

Answer 28

I like the way you worked ytour problem, @brinethery :)

Answer 29

Thanks!