Question

Solve [(D³+D²D¹-D(D¹)^2-(D¹)^3]z = e^xcos2y

Answer 1

Solution;

For the complementary solutions;

$m^3+m^2-m-1=0$

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

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

$(m+1)(m+1)(m-1)=0$

$m=-1,-1,1$

The complementary function we have;

$C.F=[x\Phi_1(y-x)+\Phi_2(y-x)]e^{-x}+\Phi_3(y+x)e^x$

The particular Integral;

$P.I=\frac{1}{f(D,D')}e^xcos2y$

$f(D,D')=D^3+D^2D'-D(D')^2-(D')^3$

From $e^xcos 2y$ Coefficient of x is 1 and

that of y is 2;

$f(1,2)=1^3+(1^2×2)-(1×2^2)-(2)^3=-9$

Hence;

$P.I=-\frac19e^xcos2y$

Hence the general solution of the equation is;

$z=[x\Phi_1(y-x)+\Phi_2(y-x)]e^{-x}+\Phi_3(y+x)e^x-\frac19e^xcos2y$