## haganmc 3 years ago how to solve this diff equation?? dx/dt−x^3=x

1. lgbasallote

have you tried Bernoulli's?

2. Algebraic!

dx/dt = x+x^3 divide by x+x^3 multiply by dt

3. haganmc

the final answer should be $x=\pm \sqrt{C ^{2t}/(1-Ce ^{2t})}$

4. ironmanjimbo

Multiplying by dt is impossible! Rates of change are as is.

5. Algebraic!

lol

6. haganmc

you get $\frac{ 1 }{ x+x^3 }dx=dt$

7. haganmc

but how do you integrate this??

8. ironmanjimbo

haganmc is right on, what he did is separation of variables which is not the same as multiplying by a component of a rate of change. Well done. Now go with factoring and partial fraction decomposition and you are on your way to a solution

9. Algebraic!

@haganmc ... looks good to me.

10. Algebraic!

11. Algebraic!

12. haganmc

no now You have to integrate both sides.. i am trying to get x by itself

13. haganmc

but how do you do partial fraction decomposition?

14. Algebraic!

use partial frac. expansion on 1/(x+x^3)

15. Algebraic!

A / x + Bx /(x^2+1)

16. haganmc

thats where i messed up... i didnt put an x after the B

17. ironmanjimbo

Algebraic, YES, nice going!

18. Algebraic!

_Multiplying by dt is impossible! Rates of change are as is._

19. ironmanjimbo

That is correct. What you are actually doing is separation of variables, It is an important distinction!!!

20. Algebraic!

go learn the chain rule

21. ironmanjimbo

I don't mind if you do not want to agree with me, but I will simply suggest that you look into what I'm pointing out regarding how the rate of change is separated in such equations. Simply put, it LOOKS like dt was multiplied, but it was not.