A community for students.
Here's the question you clicked on:
 0 viewing
RolyPoly
 3 years ago
Differential equation
Did I make any mistakes here?
\[\frac{dx}{dt}  x^3 = x\]\[\frac{dx}{dt} = x^3 + x\]\[\frac{dx}{x^3+x} = dt\]\[(\frac{1}{x} \frac{x}{x^2+1})dx = dt\]\[\ln x  tan^{1}x = t + c\]
RolyPoly
 3 years ago
Differential equation Did I make any mistakes here? \[\frac{dx}{dt}  x^3 = x\]\[\frac{dx}{dt} = x^3 + x\]\[\frac{dx}{x^3+x} = dt\]\[(\frac{1}{x} \frac{x}{x^2+1})dx = dt\]\[\ln x  tan^{1}x = t + c\]

This Question is Closed

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ dx }{ x^3+1 }=dt\] Hmm why isn't it equal to this? :o Maybe I'm missing something.\[\frac{ dx }{ x^3+x }=dt\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Ah ok i thought so :) imma check the fraction decomp a sec

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\frac{ 1 }{ x^2+1 }dx=\tan^{1}x\] \[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx \neq \tan^{1}x\] Woops! That's just another natural log I think! :o

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Oops! The problem is the partial fraction!!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1It is? Hmm I got the same thing you did...

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Not here. But in my notebook :S

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx = \frac{1}{2}\ln x^2+1 +C\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.