We have given the differential equation:
(D2+DD′+D′+1)z=5ex
The given equation cannot be written as linear factors. Hence, it's complimentary function is taken as a trial solution
z=∑Aehx+ky
Therefore we have,
Dz=∑Ahehx+ky
D′z=∑Akehx+ky
DD′z=∑Ahkehx+ky
Put all these values in the given equation (D2+DD′+D′+1)z=0 , We have
∑Ah2ehx+ky+∑Ahkehx+ky+∑Akehx+ky+∑Aehx+ky=0
∑A(h2+hk+k+1)ehx+ky=0⟹(h2+hk+k+1)=0
k=−(h+1)(h2+1)
Thus,
CF=∑Aehx+ky,wherek=−(h+1)(h2+1)
PI=F(D,D′)1eax+by
PI=5D2+DD′+D′+11ex
PI=512+1.0+0+11ex
PI=25ex
Hence, the required solution is =CF+PI
=∑Aehx+ky+25ex,wherek=−(h+1)(h2+1)
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