The actual question is;
(D2+4DD′+3(D′)2)z=ycosx,where D=∂x∂,and D′=∂y∂
To find the complementary function (C.F), let D=m,and D′=1, then the auxiliary equation of the given PDE in terms of variable m is;
m2+4m+3=0
⇒(m+1)(m+3)=0⇒m=−1,−3.
Thus the complementary function of the given PDE is;
C.F=zc=f1(y−x)+f2(y−3x)
Next, to find the particular integral (P.I) of the given PDE, we proceed as follows;
P.I=(D2+4DD′+3(D′)2)1ycosx=(D+D′)(D+3D′)1ycosx
=(D+D′)1∫(c+3x)cosx dx,where y=c+3x
=(D+D′)1[(c+3x)sinx+3cosx], by integration by parts
=(D+D′)1[ysinx+3cosx], since y=c+3x
=∫[(c+x)sinx+3cosx] dx,where we put y=c+x
=∫[(c+x)sinx dx]+3∫cosx dx
=[(c+x)(−cosx)+sinx]+3sinx ,integration by parts
=−ycosx+4sinx, where we put back y=c+x
Thus the particular integral of the given PDE is;
P.I=zp=−ycosx+4sinx
Since complete solution (z) is z=zc+zp, thus we have the complete solution as;
z=f1(y−x)+f2(y−3x)−ycosx+4sinx
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