Answer to Question #229306 in Differential Equations for Hridoy

Question

Solve it
(D^2+DD'-2D'^2)z= (y-1)e^x

Answer 1

D²+DD'-2D'²=(y-1)e^x

D=m

D'=1

m²+m-2=0

m²-m+2m-2=0

(m+2)(m-1)=0

m=-2 or 1

The roots remain distinct

C.F=f₁(y-2x)+f₂(y+x)

P.I=$\frac{1}{D^{2}+DD'-2D'^{2}}•(y-1)e^{x}$

=$\frac{1}{(D+2D')(D-D')}•(y-1)e^{x}$

=$[\frac{1}{D+2D'}][\frac{1}{D-D'}•(y-1)e^{x}]$

D-mD'=D-D'

$\therefore m=1$

y=C-mx=C-x

P.I=$\frac{1}{D+2D'}[\int (C-x-1)e^{x}dx]$

=$\frac{1}{D+2D'}[(C-x-1)e^{x}-e^{x}]$

=$\frac{1}{D+2D'}[(y-1)e^{x}-e^{x}]$

D-mD'=D+2D'

$\therefore m=-2$

y=C-mx=C+2x

P.I=$\int ((C+2x-1)e^{x}-e^{x})dx$

=(C+2x-1)e^x-(2)e^x--e^x

=(y-1)e^x--2e^x--e^x

=(y-4)e^x

Complete solution,Z=C.F+P.I

=$f_1(y-2x)+f_2(y+x)+(y-4)e^{x}$