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mathew0135
Solve the initial value problem: x(dy/dx)+y(x) = 9y(x)^(2), y(1) = -1
\[x \frac{ dy }{dx } + y(x) = 9y(x)^2 , y(1) =-1\] Just the equation a little neater
I would solve for dy/dx and intergrate.
the ex's int he brackets are just indicating the independent variable right?\[x \frac{ \text dy }{\text dx } + y = 9y^2 , \]
I believe so, that's just how its been written. I'll try solving for dy/dx and integrating then,
So I would sub those in.
u realize that you can separate the variables easily here ?
\[\frac{ dy }{ dx } = \frac{ 9y(x)^2-y }{ x }\] Not too familiar with these problems, but basically i need to integrate that, no?
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
can u integrate both sides now ?
think i got it, left an x over the y side so i confused my self. \[1 = \ln(1-9y)-\ln(y)\]
u integrated y-variable correctly , but what about \(\int (1/x)dx\) its not =1
i meant u should get something like this : \(\ln x=ln(1-9y)-lny+c\) then use logarithmic properties to simplify
Okay, catching on think i figured out what i did wrong when i integrated last anyway.
Maybe y = \[\frac{ 1 }{ x+9 }\] May be a final solution, just simplifying the equation?
ln cx= ln |(1-9y)/y| cxy= (1-9y) now use y(1) = -1 to find c.
u got this simplification ? ---->ln cx= ln |(1-9y)/y|
yup, \[C(-1)(1)=1-9(-1)\] \[-C=10\] \[C=-10\] \[10xy=(1-9y)\]
-10xy =1-9y or 10xy-9y+1=0
ahh, forgot the negative, think i have enough to try a few more of these questions any way. Thank you. :)