## anonymous 3 years 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

1. anonymous

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

2. anonymous

Proof that I actually tried hehe...

3. Luigi0210

@Jhannybean

4. anonymous

What if we expanded the RHS first?

5. anonymous

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

6. anonymous

Expand and use integrating factor

7. anonymous

Ah...

8. anonymous

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

9. anonymous

I can't do it. I get stuck.

10. anonymous

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

11. anonymous

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

12. anonymous

Yep.

13. AccessDenied

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?

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. zepdrix

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

18. zepdrix

towel throwing time? :c aw

19. anonymous

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.

20. zepdrix

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

21. AccessDenied

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. :)

22. anonymous

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.

23. AccessDenied

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

24. anonymous

That looks much better.

25. anonymous

Zepdrix was right.

26. anonymous

And accessdenied too :-). Thanks everyone!

27. AccessDenied

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

28. anonymous

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

29. anonymous

Thanks!