Question

solve the equations
d^3y/dx^3 -3d^2y/dx^2 + 4y=0​

Answer 1

Answer:

$y - 3xy + \frac{2 {x}^{3}y }{3} = \frac{ {x}^{2}c }{2} + d$

Step-by-step explanation:

$\frac{ {d}^{3 }y }{ d{x}^{3} } - 3 \frac{ {d}^{2}y}{d {x}^{2} } + 4y = 0 \\ \: integration \: on \: both \: sides \\ \frac{ {d}^{2 }y }{ d{x}^{2} } - 3 \frac{ {d}y}{d {x} } + 4yx = c \\ again \: \: integration \: on \: both \: sides \\ \frac{ {d}y }{ d{x} } - 3y + 2{x}^{2}y = cx \\ again \: \: integration \: on \: both \: sides \\ y - 3xy + \frac{2 {x}^{3}y }{3} = \frac{ {x}^{2}c }{2} + d$