## haganmc Group Title how to solve this diff equation?? dx/dt−x^3=x one year ago one year ago

1. lgbasallote Group Title

have you tried Bernoulli's?

2. Algebraic! Group Title

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

3. haganmc Group Title

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

4. ironmanjimbo Group Title

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

5. Algebraic! Group Title

lol

6. haganmc Group Title

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

7. haganmc Group Title

but how do you integrate this??

8. ironmanjimbo Group Title

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! Group Title

@haganmc ... looks good to me.

10. Algebraic! Group Title

11. Algebraic! Group Title

12. haganmc Group Title

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

13. haganmc Group Title

but how do you do partial fraction decomposition?

14. Algebraic! Group Title

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

15. Algebraic! Group Title

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

16. haganmc Group Title

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

17. ironmanjimbo Group Title

Algebraic, YES, nice going!

18. Algebraic! Group Title

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

19. ironmanjimbo Group Title

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

20. Algebraic! Group Title

go learn the chain rule

21. ironmanjimbo Group Title

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.