Question

(D+1)(D+D'-1)z=sin(2x+3y) complete solutions

Answer 1

Answer:

it's a cos sin teta Cora meta alpha beta gama

Step-by-step explanation:

because if we divide d+1 by 2 the the equation becomes a root then it should be divided with sin cos teta alpha

Answer 2

Answer:

$Z= e^{-x}\phi(y)+e^{x}\phi(y-x)-\frac{1}{130} \left( 2cos(2x+3y)+11sin(2x+3y)\right)$

Step-by-step explanation:

C.F of $(D-m_1D'-k_1)(D-m_2D'-k_2)..... =0$ is given by

$e^{k_1x}(y+m_1x)+e^{k_2x}(y+m_2x)+.........$

so the C.F of the given by is given by $e^{-x}\phi(y)+e^{x}\phi(y-x)$ .

P.I is given by,

$P.I.=\frac{1}{D^2+DD'+D'-1} sin(2x+3y)\\\\= \frac{1}{-(2)^2+(-2*3)+D'-1)} sin(2x+3y) =\frac{1}{D'-11} sin(2x+3y)\\\\=\frac{D'+11}{D'^2-121} sin(2x+3y)=\frac{D'+11}{-3^2-121} sin(2x+3y)=\frac{D'+11}{-130} sin(2x+3y)\\=-\frac{1}{130} \left( 2cos(2x+3y)+11sin(2x+3y)\right)$

final solution will be

$C.F+P.I=e^{-x}\phi(y)+e^{x}\phi(y-x)-\frac{1}{130} \left( 2cos(2x+3y)+11sin(2x+3y)\right)$