Answer to Question #192640 in Differential Equations for Jess

Question

(d^2+dd'+d'+1) =5e^x

Answer 1

We have given the differential equation:

$(D^2+DD'+D'+1)z = 5e^x$

The given equation cannot be written as linear factors. Hence, it's complimentary function is taken as a trial solution

$z = \sum Ae^{hx+ky}$

Therefore we have,

$Dz = \sum Ahe^{hx+ky}$

$D'z = \sum Ake^{hx+ky}$

$DD'z = \sum Ahke^{hx+ky}$

Put all these values in the given equation $(D^2+DD'+D'+1)z = 0$ , We have

$\sum Ah^2e^{hx+ky}+\sum Ahke^{hx+ky}+\sum Ake^{hx+ky}+\sum Ae^{hx+ky} = 0$

$\sum A(h^2+hk+k+1)e^{hx+ky} = 0 \implies (h^2+hk+k+1) = 0$

$k = -\dfrac{(h^2+1)}{(h+1)}$

Thus,

$CF = \sum Ae^{hx+ky}, where k = -\dfrac{(h^2+1)}{(h+1)}$

$PI = \dfrac{1}{F(D,D')}e^{ax+by}$

$PI = 5\dfrac{1}{D^2+DD'+D'+1}e^x$

$PI = 5\dfrac{1}{1^2+1.0+0+1}e^x$

$PI = \dfrac{5e^x}{2}$

Hence, the required solution is $= CF+PI$

$= \sum Ae^{hx+ky}+\dfrac{5e^x}{2}, where\hspace{2mm} k = -\dfrac{(h^2+1)}{(h+1)}$