Question

Differential equation
Did I make any mistakes here?
\[\frac{dx}{dt} - x^3 = x\]\[\frac{dx}{dt} = x^3 + x\]\[\frac{dx}{x^3+x} = dt\]\[(\frac{1}{x}- \frac{x}{x^2+1})dx = dt\]\[\ln |x| - tan^{-1}x = t + c\]

Answer 1

\[\frac{ dx }{ x^3+1 }=dt\]
Hmm why isn't it equal to this? :o Maybe I'm missing something.\[\frac{ dx }{ x^3+x }=dt\]

Answer 2

Sorry, it was a typo!

Answer 3

Ah ok i thought so :) imma check the fraction decomp a sec

Answer 4

\[\int\limits_{}^{}\frac{ 1 }{ x^2+1 }dx=\tan^{-1}x\]
\[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx \neq \tan^{-1}x\]
Woops! That's just another natural log I think! :o

Answer 5

Oops! The problem is the partial fraction!!

Answer 6

It is? Hmm I got the same thing you did...

Answer 7

Not here. But in my notebook :S

Answer 8

oh heh

Answer 9

\[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx = \frac{1}{2}\ln |x^2+1| +C\]

Answer 10

yay team c:

Answer 11

Thanks :)