Question

Solve the separable equation: dx/dt - x^3 = x

Answer 1

since its separable; just separate it ...

Answer 2

might have to delve into partial fractions for it

Answer 3

and because it's an equation equat it?

Answer 4

i will write up an equation, give me a minute

Answer 5

\[\int\limits_{}^{} \frac{ dx }{ dt } = \int\limits_{}^{} x+x^3\]
\[\int\limits_{}^{} \frac{ dx }{ x+x^3 } = \int\limits_{}^{} dt\]
\[-\frac{ \ln(x^2+1) }{ 2 } + \ln \left| x \right| = t\]

Answer 6

oops i forgot the + c (constant)

Answer 7

how did you go from the 2nd to 3rd equation?

Answer 8

http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^3%2Bx%29
:P lazy write equation out...

Answer 9

gonna go sleep goodnight people

Answer 10

pleasant dreams .....

Answer 11

thanks :D

Answer 12

\[\frac{1}{x+x^3}=\frac{1}{x(1+x^2)}=\frac{A}{x}+\frac{Bx+C}{1+x^2}\]multiply by \(x\)\[\frac{1}{1+x^2}=A+x\frac{Bx+C}{1+x^2}\]let \(x=0\)\[A=1\]------------------------
multiply by \(1+x^2\)\[\frac{1}{x}=\frac{A(1+x^2)}{x}+Bx+C\]let \(x=i\)\[-i=Bi+C\]\[B=-1 , C=0\]-------------------------\[\frac{1}{x+x^3}=\frac{1}{x}+\frac{-x}{1+x^2}\]

Answer 13

Thank you so much, SNSDYoona. I'm sorry it took me so long to see this. You helped me out greatly. I appreciate it.