## mathew0135 3 years ago Solve the initial value problem: x(dy/dx)+y(x) = 9y(x)^(2), y(1) = -1

1. mathew0135

$x \frac{ dy }{dx } + y(x) = 9y(x)^2 , y(1) =-1$ Just the equation a little neater

2. Dido525

I would solve for dy/dx and intergrate.

3. UnkleRhaukus

the ex's int he brackets are just indicating the independent variable right?$x \frac{ \text dy }{\text dx } + y = 9y^2 ,$

4. Dido525

and we know y(1) = -1.

5. mathew0135

I believe so, that's just how its been written. I'll try solving for dy/dx and integrating then,

6. Dido525

So I would sub those in.

7. hartnn

u realize that you can separate the variables easily here ?

8. mathew0135

$\frac{ dy }{ dx } = \frac{ 9y(x)^2-y }{ x }$ Not too familiar with these problems, but basically i need to integrate that, no?

9. hartnn

u bring all terms of one variable on one side of = sign, like this : $$\large \frac{1}{9y^2-y}dy=\frac{1}{x}dx$$ then integrate

10. hartnn

can u integrate both sides now ?

11. mathew0135

doing that now

12. hartnn

13. mathew0135

think i got it, left an x over the y side so i confused my self. $1 = \ln(1-9y)-\ln(y)$

14. hartnn

u integrated y-variable correctly , but what about $$\int (1/x)dx$$ its not =1

15. hartnn

i meant u should get something like this : $$\ln x=ln(1-9y)-lny+c$$ then use logarithmic properties to simplify

16. mathew0135

Okay, catching on think i figured out what i did wrong when i integrated last anyway.

17. mathew0135

Maybe y = $\frac{ 1 }{ x+9 }$ May be a final solution, just simplifying the equation?

18. hartnn

ln cx= ln |(1-9y)/y| cxy= (1-9y) now use y(1) = -1 to find c.

19. hartnn

u got this simplification ? ---->ln cx= ln |(1-9y)/y|

20. mathew0135

yup, $C(-1)(1)=1-9(-1)$ $-C=10$ $C=-10$ $10xy=(1-9y)$

21. hartnn

-10xy =1-9y or 10xy-9y+1=0

22. mathew0135

ahh, forgot the negative, think i have enough to try a few more of these questions any way. Thank you. :)

23. hartnn

welcome ^_^