## anonymous one year ago Differential Equation Problem: How Can i solve the Initial value problem. I am not sure if to use integrating factor? Equation below.

1. anonymous

$2y \prime -y^2 + 4 =0, y(0) = 0$

2. amistre64

what methods are available to you?

3. anonymous

integrating factory and separable....

4. amistre64

2 dy/dx = y^2-4 looks separable to me ... not sure if it makes life easier tho

5. anonymous

Yep, i can work with that.

6. anonymous

I'll post the result soon.

7. anonymous

It does not. It is complex.

8. anonymous

|dw:1443368763254:dw|

9. Loser66

Use partial fraction.

10. Loser66

|dw:1443368964936:dw|

11. anonymous

I'm using that, A = -1/2 and b = 1/2

12. anonymous

is it safe to say....|dw:1443369358723:dw|

13. anonymous

but as y (0) = 0. I noticed that the equation initial condition will bring take the solution to.... e^(2c) = -1, but c does not exist since the ln(-1) is undefined.

14. amistre64

as per the wolf ... might help you see where to go.

15. amistre64

A and B are +- 1/4 right?

16. amistre64

i ignored the 2 lol

17. amistre64

dy/(y^2-4) = dx/2 ln((y-2)/(y+2))/4 = x/2 +C ln((y-2)/(y+2)) = 2x +4C (y-2)/(y+2) = e^(2x+4C) = C e^(2x) x=0,y=0 -1 = C

18. amistre64

y-2 = -e^(2x) (y+2) y+ye^(2x) = -2e^(2x)+2 y(1+e^(2x)) = -2(1-e^(2x)) ...

19. amistre64

mighta factored the rhs off by a negative in my head

20. anonymous

Thanks all.