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anonymous
 5 years ago
Solve the differential equation.
xy'  (x + 1)y = 0
anonymous
 5 years ago
Solve the differential equation. xy'  (x + 1)y = 0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x(dy/dx)  (x+1)y = 0 x(dy/dx) = (x+1)y dy/dx = [ (x+1)/x ] y 1/y (dy) = (x+1)/x dx ::: integrate can yu go from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0somehow i got (x1)/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0noo instead of (x+1)/x dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my final answer was y = 1/2Ce^x(x)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a feeling that's not right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm not what i got, wait i'm gonna type what i have

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok now i'm up to ln(y) = x + ln(x) + C.. not sure what to do next

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[xy'  (x + 1)y = 0\]\[x{dy \over dx}  (x+1)y = 0\]\[x{dy \over dx} = (x+1)y\]\[{dy \over dx} = {(x+1) \over x} (y)\]\[{1 \over y} dy = {(x+1) \over x} dx\] Integrate:\[\lny= \lnx+x +C\]\[e^{lny} = e^{lnx+x+C}=e^{lnx}*e^{x}*e^{C}\]\[y = A*x*e^{x}\] \[y = Axe^{x}\] that's what i have

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think soo.. what is A?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0e^C = constant a constant which i called A, you can called it whatever..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but did yu understand everything?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh gotcha.. i understand it.. thank you.. can you help me with other problems too?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good. i sure can, just post them and i'll try

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1/2}^{2} \] [(e^1/x) / (x^2)] dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{2}{{e^{1} \over x} \over x^{2}} dx\] is that what yu tried to write

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{e^{1/x} \over x^{2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that didn't come out on my computer  maybe bc i'm using a mac. but its the integration bounded from 1/2 to 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the function that you just wrote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do a product rule between e^(1/x) and x^(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not really sure.. the integration of e^(1/x) is just e^(1/x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer comes up to be e^2  sqrt(e) i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that?
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