Question

求微分方程 y的二阶导数+y的一阶导数=X的平方 的通解... 求详细求解过程~

y’= ax3 + bx2 + cx + d
为什么是令特解成这个形式？

Answer 1

d²y/dx² + dy/dx = x²

Solution :

d²y/dx² + dy/dx = 0

The characteristic equation (特征方程) is
λ² + λ = 0， λ(λ + 1) = 0, λ₁= 0, λ₂= -1.

So the characteristic solution (特征解) is
y = c₁+ c₂e^(-x)

Let particular solution (令特解) be
y’= ax³ + bx² + cx + d

Substituting y' into the differential equation (将特解代入原方程)：
6ax + 2b + 3ax² + 2bx + c = x²
a = 1/3, b = -3a = -1, c = -2b = 2

So，the general solution (通解) is
y = c₁+ c₂e^(-x) + (1/3)x³ - x² + 2x

补充说明:
因为方程的右面是二次代数式,左边至少求导一次,特解一定是三次式,求导后才有二次代数式.