Differential Calculus

if the given

mass= 590 kg
depth 50 meters

find the general solution

2y ln y dx - x (1 + 2x ln y) dy = 0

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?