Answer to Question #285372 in Differential Equations for Abu bakar

Question

Solve

(D^(2)-DD'-2D'^(2)+2D+2D')z=sin(2x+y)

Answer 1

$(D^2-DD'-2D'^2+2D+2D')z=sin(2x+y)$

auxillary equation:

$m^2-m-2+2m+2=0$

$m^2+m=0$

$m_1=0,m_2=-1$

complementary function:

$C.F.=f_1(y+m_1x)+f_2(y+m_2x)=f_1(y)+f_2(y-x)$

 particular integral:

$P.I.=\frac{1}{F(D,D')}sin(ax+by)=I.P.\frac{1}{F(D,D')}e^{i(ax+by)}=I.P.\frac{1}{F(ia,ib)}e^{i(ax+by)}$

$P.I.=I.P.\frac{1}{(2i)^2-2i\cdot i-2i^2+2\cdot2i+2i}e^{i(ax+by)}=I.P.\frac{1}{6i}(cos(2x+y)+isin(2x+y))=$

$=I.P.-\frac{i}{6}(cos(2x+y)+isin(2x+y))=-\frac{1}{6}cos(2x+y)$

$z=C.F.+P.I.=f_1(y)+f_2(y-x)-\frac{1}{6}cos(2x+y)$