## stottrupbailey Group Title calc II separable equation, please help :) one year ago one year ago

1. stottrupbailey Group Title

$\frac{ du }{ dt }=e ^{6u+8t}$ given that u(0)=8. So this is how I set it up $\int\limits_{}^{}e ^{-6u}du=\int\limits_{}^{}e ^{8t}dt$

2. stottrupbailey Group Title

Is that right so far?

3. nubeer Group Title

yes looks right. so far.

4. stottrupbailey Group Title

Okay, so this is what I got for my integration: $\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }+c$

5. stottrupbailey Group Title

and then using that u(0)=8, I found that c=-0.125.

6. stottrupbailey Group Title

I actually have already worked through this problem, but I got the wrong answer at the end so I'm trying to figure out where I'm going wrong

7. stottrupbailey Group Title

so then I end up with $\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }-0.125$

8. stottrupbailey Group Title

so then my final answer was u(t)=$\frac{ 1 }{ 6 }\ln(\frac{ 3 }{ 4 }e ^{8t}-\frac{ 3 }{ 4 })$

9. stottrupbailey Group Title

but the book says that isn't the right answer

10. nubeer Group Title

hmm well i also got c same as u.. well what answer book says?

11. stottrupbailey Group Title

well, it's actually webwork, so it just tells me that my answer is wrong, I'm not sure what the right answer is

12. stottrupbailey Group Title

I'm thinking something must be wrong with the way I'm doing the algebra to rearrange it in terms of u(t)?

13. nubeer Group Title

well i am not sure but the way u solved question looks fine and the answer should be the one u got..

14. stottrupbailey Group Title

hmmm. Okay. I'll ask the tutor at 8. Thanks anyway :)