Question

obtain the differential equation y=ax^3+bx^2

Answer 1

y = ax³ + bx²
differentiate wrt x
dy/dx = 3ax² + 2bx -----(1)

again differentiate wrt x
d²y/dx² = 6ax + 2bx-----(2)

from equations (1) and (2)
we get
ax = 1/3{ d²y/dx² - 1/x.dy/dx}
b = 1/2{2.1/x.dy/dx -d²y/dx²}
put it in given equation

y = 1/3{d²y/dx² -1/x.dy/dx}x² + 1/2{2.1/x.dy/dx -d²y/dx²}x²

6y = 2x²d²y/dx² - 2x.dy/dx +3x.dy/dx - 3x².d²y/dx²

6y = x.dy/dx -x².d²y/dx²