Y '' + y = x ^ 2 differential equation Y '' + y = x ^ 2 differential equation

Y '' + y = x ^ 2 differential equation Y '' + y = x ^ 2 differential equation

The corresponding characteristic equation of Y & # 39; &# 39; + y = 0 is R ^ 2 + 1 = 0, and its solution is: r = ± I, so the general solution of Y & # 39; &# 39; + y = 0 is: y = C1 · cosx + C2 · SiNx. A special solution can be set as: y * = ax ^ 2 + BX + C. substituting the solution of the original equation into: 2A + ax ^ 2 + BX + C = x ^ 2, the solution is: a = 1, B = 0, C = - 2, so a special solution is: y * =

Find the general solution of the differential equation y ″ + y ′ = x2

The corresponding characteristic equation of homogeneous equation is λ 2 + λ = 0, the characteristic root is λ 1 = 0, and λ 2 = - 1. Therefore, the general solution of homogeneous equation is Y1 = C1 + C2 e-x. because the non-homogeneous term is f (x) = x2 = x2e0, and 0 is the single characteristic root, the general solution of the original equation can be assumed to be y * = x (AX2 + BX + C). Substituting into the original equation, a = 13, B = - 1, C = 2, so The general solution of the original equation is y = Y1 + y * = C1 + C2 E-X + 13x3-x2 + 2x

General solution of differential equation y '' + y '/ (1-x) = 0

y''+y'/(1-x)=0
y''=y'/(x-1)
The solution is y '= C (x-1)
Points:
y=C(x-1)²/2+C2
=C1(x-1)²+C2