Answer to Question #275244 in Differential Equations for Zari

Question

Solve differential equation

(D^3-6D’D+11D’^2D-6D’)z=cos(2x+y)+e^(2x+y)-y

Answer 1

$(D^3-6D’D+11D’^2D-6D’)z=cos(2x+y)+e^{2x+y}-y$

auxillary equation:

$m^3-6m+11m-6=m^3+5m-6=0$

$(m-1)(m^2+m+6)=0$

complementary function:

$C.F.=f_1(y+x)+\sum c_re^{a_rx+b_ry}$

where

$a_r^2+b_r+6=0$

then:

$C.F.=f_1(y+x)+\sum c_re^{a_rx-(6+a_r^2)y}=f_1(y+x)+e^{-6y}\sum c_re^{a_r(x-a_ry)}$

particular Integral:

$P.I._1=\frac{e^{2x+y}}{D^3-6D’D+11D’^2D-6D’}$ (replace D by 2 and D' by 1)

$=\frac{e^{2x+y}}{8-12+22-6}=\frac{e^{2x+y}}{12}$

$P.I._2=\frac{cos(2x+y)}{D^3-6D’D+11D’^2D-6D’}$ (replace D²=-a²=-4,DD'=-ab=-2,D'²=-b²=-1)

$=\frac{cos(2x+y)}{-4D+12-11D-6D'}=\frac{cos(2x+y)}{-15D+12-6D'}=-\frac{cos(2x+y)/15}{D+0.4D'-0.8}$

$\frac{1}{D-mD'}F(x,y)=\int F(x,c-mx)dx$

where c is replaced by y+mx after integration

then:

$P.I._2=-\frac{cos(2x+y)/15}{D+0.4D'-0.8}=-\frac{1}{15}\intop cos(2x+c+0.4x)dx$

$=-\frac{sin(2.4x+c)}{15\cdot2.4}=-\frac{sin(2x+y)}{36}$

$P.I._3=\frac{-y}{D^3-6D’D+11D’^2D-6D’}=-\frac{1}{D^3}(1-\frac{6D'}{D^2}+\frac{11D'^2}{D^2}-\frac{D'}{d^3})^{-1}y=$

$=-\frac{1}{D^3}(1+\frac{6D'}{D^2}-...)y=-\frac{1}{D^3}(y+\frac{6}{D^2})=-\frac{1}{D^3}(y+3x^2)=-(yx^3/3+x^5/20)$

$z=C.F.+P.I.=$

$=f_1(y+x)+e^{-6y}\sum c_re^{a_r(x-a_ry)}+\frac{e^{2x+y}}{12}-\frac{sin(2x+y)}{36}-(yx^3/3+x^5/20)$