Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Need help with separable differential equation... I keep getting the wrong answer :/

OpenStudy (unklerhaukus):

i can help

OpenStudy (lgbasallote):

i can watch @UnkleRhaukus do his thing

OpenStudy (anonymous):

ok I just uploaded the question. The last one I did was right, but this one is incorrect. I was just wondering if someone could point out what I'm doing wrong. I'll post my work too.

OpenStudy (unklerhaukus):

\[\frac{\text dy}{\text dx}=0.6(y-200)\] \[\frac{\text dy}{y-200}=0.6\text dx\] \[\int\frac{\text dy}{y-200}=0.6\int\text dx\]

OpenStudy (anonymous):

That is what I did... maybe I am taking the integral wrong. On the left side I got ln(y-200) and on the right I got 0.6x+c

OpenStudy (unklerhaukus):

that is right

OpenStudy (anonymous):

ok then after that I plugged in and solved for C. I ended up getting ln(145). Is that correct too? :/

OpenStudy (anonymous):

anyway, after I did some work, my final answer was y=e^(0.6t)+345. Maybe my algebra is rusty or I am missing something.

OpenStudy (unklerhaukus):

\[\ln(y-200)=0.6t+c\]\[y-200=ke^{0.6t}\]\[y=ke^{0.6t}+200\] \[y(0)=55\]\[55=ke^{0.6\times0}+200\]\[55=k+200\]\[k=-145\] \[y=-145e^{0.6t}+200\]

OpenStudy (anonymous):

I'm sorry if this is a stupid question, but where did the K come from?

OpenStudy (anonymous):

also, thank you for your help. The answer was correct, but I just want to make sure I know how to do it.

OpenStudy (unklerhaukus):

\[\ln(y-200)=0.6t+c\] \[e^{\ln(y-200)}=e^{0.6t+c}\] \[y-200=e^{0.6t}e^c,\qquad\qquad \text{letting} \quad e^c=k\] \[y-200=ke^{0.6t}\]

OpenStudy (anonymous):

thank you mr. Rhaukus. You are really awesome.

OpenStudy (unklerhaukus):

:D

OpenStudy (anonymous):

man I'm sorry, but I thought I finally got this down :(. I got stuck again. I'm gonna just write it out by hand. I did this question the same way you did, but the answer is incorrect. I'm so disappointed :/. I wish I could just get this down already :/

OpenStudy (unklerhaukus):

\[\frac{\text dy}{\text dx}+6y=5\]\[\int\frac{1}{5-6y}\text dy=\int \text dx\]\[\frac{-1}{6}\int\frac{-6}{5-6y}\text dy=\int \text dx\]\[\frac{-1}{6}\ln|5-6y|=x+c\]

OpenStudy (unklerhaukus):

the rest of your work looks good

OpenStudy (anonymous):

ahhhhhh I feel so dumb xD. I forgot to do the u substitution when I took the integral xD..... Why do I suck at math? haha xD

OpenStudy (anonymous):

I think I'm way too careless.

OpenStudy (unklerhaukus):

dont feel dumb , DE are hard

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!