Question

Solve the differential equation:
(2x+3) + (2y-2)y' = 0

I know it's exact but not sure how to solve. Thanks!

Answer 1

As my answer, I've gotten y^2-2y= c, but i dont think that's right

Answer 2

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.

Answer 3

Thanks a lot!

Answer 4

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....

Answer 5

IGNORE ALL OF THAT I MEANT TO POST SOMEWHERE ELSE