Answer to Question #192640 in Differential Equations for Jess

Question #192640

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


1
Expert's answer
2021-05-17T02:49:01-0400

We have given the differential equation:


(D2+DD+D+1)z=5ex(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=Aehx+kyz = \sum Ae^{hx+ky}


Therefore we have,


Dz=Ahehx+kyDz = \sum Ahe^{hx+ky}


Dz=Akehx+kyD'z = \sum Ake^{hx+ky}


DDz=Ahkehx+kyDD'z = \sum Ahke^{hx+ky}


Put all these values in the given equation (D2+DD+D+1)z=0(D^2+DD'+D'+1)z = 0 , We have


Ah2ehx+ky+Ahkehx+ky+Akehx+ky+Aehx+ky=0\sum Ah^2e^{hx+ky}+\sum Ahke^{hx+ky}+\sum Ake^{hx+ky}+\sum Ae^{hx+ky} = 0


A(h2+hk+k+1)ehx+ky=0    (h2+hk+k+1)=0\sum A(h^2+hk+k+1)e^{hx+ky} = 0 \implies (h^2+hk+k+1) = 0


k=(h2+1)(h+1)k = -\dfrac{(h^2+1)}{(h+1)}


Thus,


CF=Aehx+ky,wherek=(h2+1)(h+1)CF = \sum Ae^{hx+ky}, where k = -\dfrac{(h^2+1)}{(h+1)}


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


PI=51D2+DD+D+1exPI = 5\dfrac{1}{D^2+DD'+D'+1}e^x


PI=5112+1.0+0+1exPI = 5\dfrac{1}{1^2+1.0+0+1}e^x


PI=5ex2PI = \dfrac{5e^x}{2}


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


=Aehx+ky+5ex2,wherek=(h2+1)(h+1)= \sum Ae^{hx+ky}+\dfrac{5e^x}{2}, where\hspace{2mm} k = -\dfrac{(h^2+1)}{(h+1)}











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