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

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zepdrix
 4 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.\[\frac{ dx }{ x^3+x }=dt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, it was a typo!

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

zepdrix
 4 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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ x }{ x^2+1 }dx = \frac{1}{2}\ln x^2+1 +C\]
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