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anonymous
 5 years ago
Solve the differential equation:
(2x+3) + (2y2)y' = 0
I know it's exact but not sure how to solve. Thanks!
anonymous
 5 years ago
Solve the differential equation: (2x+3) + (2y2)y' = 0 I know it's exact but not sure how to solve. Thanks!

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As my answer, I've gotten y^22y= c, but i dont think that's right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since this is a 1st order ODE, you can solve by separation of variables... (2y2)dy = (2x+3)dx Integrate both sides: y^2  2y = x^2 3x + C So: x^2 + y^2 +3x  2y = C... This is the general solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integration by parts: integral(uv) = uv  integral(vdu) u=arccos(x) du =  1 / √(1  x^2) dv = dx v = x Therefore: uv = arccos(x) integral(vdu) = x/√(1x^2) let k = (1x²) then dk = 2x dx and dk/2 = x dx ∫x/(1x²) * dx = (1/2)∫dk/√k = √k Therefore ignore that....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0IGNORE ALL OF THAT I MEANT TO POST SOMEWHERE ELSE
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