Differential Calculus

find a general solution

dx/dy =[ -2x (csc 2y – 2x³ cos 2y)] / sec 2y

(xy + (y/x) + (3/x) -2) dy + y² dx = 2(1+(y/x))dx

[(1-xy)^-2]dx + [y^2-x^2(1-xy)^-2]dy=0 ;when x=1, y=1

(1+x) (y' + y²)-y=0

( 2x³ - y⁴) dx + xy³ dy= 0

2xy y' = y^2 - 2 x^2

y' = y - x y^3 e^ -2 x

Please help. I need it asap.

Solve the linear differential equation:
(D5–12D4+52D3–112D2+192D–256)y = 0

When I factored it out. I got (D-4)^3 (D^2 +4)

With D-4..

My equation will be Y= (C1 + C2x + C3x^2)e^4x

But how about my D^2 + 4?

How can I add that in my equation?

A parachutist is falling with speed 176 ft/s when his parachute opens. If the air resistance is Wv2/225 lb where W is the total weight of the man and the parachute and v is in ft/s, find his speed as a function of time after the parachute opens.

If I use the suvat equations, the given resistance is the value of my what? Or I can't use the suvat equations at all?

If I use Fresultant=Fapplied-Fresistance. I can substitute the given resistance, but what would be the value or formula for getting my Fresulant and Fapplied?

Orthogonal Trajectories
y^3 - 3 = ln |Cx|

Please check if the answer I solved is correct.
My answer: y(3x^2 + c)=2