Answer to Question #229306 in Differential Equations for Hridoy

Question #229306
Solve it
(D^2+DD'-2D'^2)z= (y-1)e^x
1
Expert's answer
2021-08-25T15:01:36-0400

D2+DD'-2D'2=(y-1)ex

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=1D2+DD2D2(y1)ex\frac{1}{D^{2}+DD'-2D'^{2}}•(y-1)e^{x}

=1(D+2D)(DD)(y1)ex\frac{1}{(D+2D')(D-D')}•(y-1)e^{x}

=[1D+2D][1DD(y1)ex][\frac{1}{D+2D'}][\frac{1}{D-D'}•(y-1)e^{x}]

D-mD'=D-D'

m=1\therefore m=1

y=C-mx=C-x

P.I=1D+2D[(Cx1)exdx]\frac{1}{D+2D'}[\int (C-x-1)e^{x}dx]

=1D+2D[(Cx1)exex]\frac{1}{D+2D'}[(C-x-1)e^{x}-e^{x}]

=1D+2D[(y1)exex]\frac{1}{D+2D'}[(y-1)e^{x}-e^{x}]

D-mD'=D+2D'

m=2\therefore m=-2

y=C-mx=C+2x

P.I=((C+2x1)exex)dx\int ((C+2x-1)e^{x}-e^{x})dx

=(C+2x-1)ex-(2)ex--ex

=(y-1)ex--2ex--ex

=(y-4)ex

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

=f1(y2x)+f2(y+x)+(y4)exf_1(y-2x)+f_2(y+x)+(y-4)e^{x}


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