Question

Solve the initial value problem:

x(dy/dx)+y(x) = 9y(x)^(2), y(1) = -1

Answer 1

\[x \frac{ dy }{dx } + y(x) = 9y(x)^2 , y(1) =-1\]

Just the equation a little neater

Answer 2

I would solve for dy/dx and intergrate.

Answer 3

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

Answer 4

and we know y(1) = -1.

Answer 5

I believe so, that's just how its been written.

I'll try solving for dy/dx and integrating then,

Answer 6

So I would sub those in.

Answer 7

u realize that you can separate the variables easily here ?

Answer 8

\[\frac{ dy }{ dx } = \frac{ 9y(x)^2-y }{ x }\]

Not too familiar with these problems, but basically i need to integrate that, no?

Answer 9

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

Answer 10

can u integrate both sides now ?

Answer 11

doing that now

Answer 12

take your time :)

Answer 13

think i got it, left an x over the y side so i confused my self.

\[1 = \ln(1-9y)-\ln(y)\]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

yup, \[C(-1)(1)=1-9(-1)\]
\[-C=10\]
\[C=-10\]

\[10xy=(1-9y)\]

Answer 21

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

Answer 22

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

Answer 23

welcome ^_^