Auxillaryequation/y'' - 4y' +4

2013-03-12 2:43 am
如題 :solve y'' - 4y' +4y =2x =6(e^3x)

given y(0) =6 , y'(0) =18

回答 (1)

2013-03-12 5:58 pm
✔ 最佳答案
Solve y"-4y'+4y=2x+6(e^3x)Given y(0)=6, y'(0)=18輔助方程式: 0=m^2-4m+4=(m-2)^2 => 2, 2齊次解: y1(x)=(a+bx)e^(2x)特殊解: y2(x)=ax+b+c*e^(3x)4*y2(x)=4ax+4b+4c*e^(3x)-4*y2'=-4a-4*3c*e^(3x)y2"=9c*e^(3x)Add: 2x+6*e^(3x)=c*e^(3x)+4ax-4(b-a)a=1/2, b=a=1/2, c=6So y2(x)=2x+2+6*e^(3x)通解: y(x)=(a+bx)e^(2x)+2x+2+6*e^(3x)邊界條件: y(0)=6=a+2+6 => a=-2y'(x)=(2a+2bx+b)e^(2x)+2+18e^(3x)y'(0)=18=(2a+b)+2+18 => b=2最後解: y(x)=2(x-1)*e^(2x)+6*e^(3x)+2(x+1).........ans


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