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

Mathematics
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As my answer, I've gotten y^2-2y= c, but i dont think that's right
Since this is a 1st order ODE, you can solve by separation of variables... (2y-2)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.
Thanks a lot!

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Other answers:

Integration 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/√(1-x^2) let k = (1-x²) then dk = -2x dx and -dk/2 = x dx ∫x/(1-x²) * dx = (-1/2)∫dk/√k = -√k Therefore ignore that....
IGNORE ALL OF THAT I MEANT TO POST SOMEWHERE ELSE

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