Question

微分方程2y''-y'-y=2x^2 -3 当x=0,y'=4,y=0求特解

微分方程2y''-y'-y=2x^2 -3y=0且当x=0,y'=4求特解

Answer 1

如图

Answer 2

设y=f(x)，F(s)=L[f(x)]
两边取拉普拉斯变换，
2[s²F(s)-sf(0)-f'(0)]-[sF(s)-f(0)]-F(s)=4/s³-3/s
2s²F(s)-8-sF(s)-F(s)=(4-3s²)/s³
F(s)*(2s²-s-1)=(8s³-3s²+4)/s³
F(s)=(8s³-3s²+4)/s³(2s+1)(s-1)
=6/(s+1/2)+3/(s-1)-9/s+4/s²-4/s³
所以f(x)=6e^(-x/2)+3e^x-9+4x-2x²
检验：f'(x)=-3e^(-x/2)+3e^x+4-4x
f''(x)=3/2*e^(-x/2)+3e^x-4
左边=2f''(x)-f'(x)-f(x)
=3e^(-x/2)+6e^x-8+3e^(-x/2)-3e^x-4+4x-6e^(-x/2)-3e^x+9-4x+2x²
=2x²-3=右边