Differential Equation: Arcsecant

Prove that the function on the right-hand is a solution of the differential equation on the left-hand:

right hand:(1+x^2)y' = xy left hand:y = sqrt 1+x^2

Integration gives the following for the arcsecant function:

int1/|x|sqrt 1-x^2 dx = arcsec x+C

Choosing parts for u and du:

u = arcsec x
du = dx

du = 1/|x|sqrt x^2 -1 dx
v = x

This is where the confusion comes in and that is what to apply to (1+x^2)y' = xy as far as do I use the Power rule of derivatives or Integration on this part and add it to y = x arcsec x - int(x*dx/sqrt(1-x^2)) during integration of parts ?

I think you have your left side and right side confused. Check out my solution to your problem. If you have any questions, feel free to reply back.

(1 + x^2) * y' = xy
y' / y = x / (1 + x^2)

u = 1 + x^2
du = 2x

y'/y = (1/2) * du / u
ln(y) = (1/2) * ln(u) + C
y = C * e^((1/2) * ln(1 + x^2))
y = C * sqrt(1 + x^2)

Is there a reason to involve the arcsecant?