1)
x = 1/t
...
x^5 d^2z/dx^2 + (2x^4 - 5x^3) dz/dx +4xz = 6x +3 can be reduced to....
d^2z/dt^2 + 5 dz/dt + 4z = 3t+6
2)
use the substitution y=x^2 to show that...
x d^2x/dt^2 + (dx/dt)^2 + 5x(dx/dt) + 3x^2 = sin2t + 3cos2t
can be converted to
d^2y/dt^2 + 5dx/dt + 6y = 2sin2t + 6cos2t
hence solce for x in terms of t
assume a solution in the form of
y=AsinBt+CcosDt then
y'= BAcosBt -CDsinDt
y"=-B^2 A sinBT + CD^2 cosDt
put those in the equation..
-B^2 A sinBT + CD^2 cosDt +5BAcosBt -6CDsinDt +6AsinBt+6CcosDt =2sin2t + 6cos2t
work with the sin terms first:
-B^2 A sinBT -6CDsinDt +6AsinBt =2sin2t
It is clear B=D=2
-4A-12C+12=2 **1
then the cosine terms...
CD^2 cosDt +5BAcosBt +6CcosDt = + 6cos2t
again, D=B=2
4C+10A+6C=6***2
now, you have two equations, two unknowns, solve for A and C. ...and you have the solution