求方程y``+y=2x²-4的通解
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y''+y=2x^2-4
The aux. equation
r^2+1=0
r=i or -i
let
yg=Acosx +Bsinx
yp=Cx^2+Dx+E
yp'=2Cx+D
yp''=2C
yp''+yp=2x^2-4
2C +(Cx^2+Dx+E) =2x^2-4
Cx^2 +Dx +(2C+E)=2x^2-4
coef. of x^2, =>C=2
coef. of x, =>D=0
coef. of constant
2C+E=-4
4+E=-4
E=-8
yp=2x^2-8
通解
y= yg+yp=Acosx +Bsinx +2x^2-8
The aux. equation
r^2+1=0
r=i or -i
let
yg=Acosx +Bsinx
yp=Cx^2+Dx+E
yp'=2Cx+D
yp''=2C
yp''+yp=2x^2-4
2C +(Cx^2+Dx+E) =2x^2-4
Cx^2 +Dx +(2C+E)=2x^2-4
coef. of x^2, =>C=2
coef. of x, =>D=0
coef. of constant
2C+E=-4
4+E=-4
E=-8
yp=2x^2-8
通解
y= yg+yp=Acosx +Bsinx +2x^2-8
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