anonymous
  • anonymous
Solve the initial value problem: x(dy/dx)+y(x) = 9y(x)^(2), y(1) = -1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[x \frac{ dy }{dx } + y(x) = 9y(x)^2 , y(1) =-1\] Just the equation a little neater
anonymous
  • anonymous
I would solve for dy/dx and intergrate.
UnkleRhaukus
  • UnkleRhaukus
the ex's int he brackets are just indicating the independent variable right?\[x \frac{ \text dy }{\text dx } + y = 9y^2 , \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
and we know y(1) = -1.
anonymous
  • anonymous
I believe so, that's just how its been written. I'll try solving for dy/dx and integrating then,
anonymous
  • anonymous
So I would sub those in.
hartnn
  • hartnn
u realize that you can separate the variables easily here ?
anonymous
  • anonymous
\[\frac{ dy }{ dx } = \frac{ 9y(x)^2-y }{ x }\] Not too familiar with these problems, but basically i need to integrate that, no?
hartnn
  • 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
hartnn
  • hartnn
can u integrate both sides now ?
anonymous
  • anonymous
doing that now
hartnn
  • hartnn
take your time :)
anonymous
  • anonymous
think i got it, left an x over the y side so i confused my self. \[1 = \ln(1-9y)-\ln(y)\]
hartnn
  • hartnn
u integrated y-variable correctly , but what about \(\int (1/x)dx\) its not =1
hartnn
  • hartnn
i meant u should get something like this : \(\ln x=ln(1-9y)-lny+c\) then use logarithmic properties to simplify
anonymous
  • anonymous
Okay, catching on think i figured out what i did wrong when i integrated last anyway.
anonymous
  • anonymous
Maybe y = \[\frac{ 1 }{ x+9 }\] May be a final solution, just simplifying the equation?
hartnn
  • hartnn
ln cx= ln |(1-9y)/y| cxy= (1-9y) now use y(1) = -1 to find c.
hartnn
  • hartnn
u got this simplification ? ---->ln cx= ln |(1-9y)/y|
anonymous
  • anonymous
yup, \[C(-1)(1)=1-9(-1)\] \[-C=10\] \[C=-10\] \[10xy=(1-9y)\]
hartnn
  • hartnn
-10xy =1-9y or 10xy-9y+1=0
anonymous
  • anonymous
ahh, forgot the negative, think i have enough to try a few more of these questions any way. Thank you. :)
hartnn
  • hartnn
welcome ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.