Question

微分方程y''=(α+β)y'+e^ (α+β)x的通解？

Answer 1

y''=(α+β)y'+e^ (α+β)x
y''-(α+β)y'=e^ (α+β)x
The aux. equation
p^2-(α+β)p=0
p(p-(α+β)) =0
p=0 or α+β
let
yg= Ae^[(α+β)x] +B
yp=Cxe^[(α+β)x]
yp'=C[(α+β)x + 1]e^[(α+β)x]
yp''
=C{ (α+β)[(α+β)x + 1] +(α+β) } e^[(α+β)x]
=C[(α+β)^2.x + 2(α+β)] e^[(α+β)x]
yp''-(α+β)yp'=e^[(α+β)x]
C[(α+β)^2.x + 2(α+β)] e^[(α+β)x] -(α+β)C[(α+β)x + 1]e^[(α+β)x] =e^[(α+β)x]
C(α+β) e^[(α+β)x=e^[(α+β)x]
C= 1/(α+β)
yp= [1/(α+β)]xe^[(α+β)x]
通解
y=yg+yp=Ae^[(α+β)x] +B +[1/(α+β)]xe^[(α+β)x]