Question

solve(d^2+dd'-6d'^2)z=cos(2x+y)

Answer 1

The auxiliary equation is; $m^{2}+m-6=0\newline (m-2)(m+3)=0\newline m=2,\ -3$

C.F = f₁(y+2x)+f₂(y-3x)

P.I = $\frac{1}{D^{2}+DD^{I}-6{D^{I}}^{2}}cos(2x+y)$

=R.P$\frac{1}{(D-2D^{I})(D+3D^{I})}e^{i(2x+y)}$

=R.P$\frac{1}{(2i+3(i))}e^{i(2x+y)}$

=R.P$\frac{-i}{5}x(cos(2x+y)+i\ sin(2x+y))$

=$\frac{-x}{5}cos(2x+y)$

z=f₁(y+2x)+f₂(y-3x)-$\frac{x}{5}cos(2x+y)$