Question

大一高数 求下列微分方程通解

Answer 1

y''+y'=2x^2.e^x
The aux. equation
p^2 +p=0
p(p+1)=0
p=0 or -1
yg = Ae^(-x) +B
yp= (Cx^2 +Dx+F)e^x
yp'=(Cx^2 +Dx+F + 2Cx+D)e^x =[Cx^2 +(2C+D)x +D+F]e^x
yp''=[Cx^2 +(2C+D)x +D+F + 2Cx + 2C+D]e^x =[Cx^2 +(4C+D)x + 2C+2D+F].e^x
yp''+yp'=2x^2.e^x
[Cx^2 +(4C+D)x + 2C+2D+F].e^x +[Cx^2 +(2C+D)x +D+F]e^x =2x^2.e^x
[ 2Cx^2 +(6C+2D)x + 2C+3D+2F] e^x =2x^2.e^x
=>
2C=2 and 6C+2D=0 and 2C+3D +2F=0
=>
C=1 and D=-3 and F=7/2
yp = (x^2 -3x+ 7/2).e^x
y''+y'=2x^2.e^x 的通解
y=yg+yp =Ae^(-x) +B +(x^2 -3x+7/2).e^x

Answer 2

这样子。。。。