Question

Solve

(D^(2)+DD'-6D'^(2))z=y cosx

Answer 1

Answer:

Given:

$$(D ^ 2 +DD ^′ −6D ^{′2} )z=ycosx$$

Now, the Auxiliary equation will be:

$m ^2 +m−6=0 ⟹(m−2)(m+3)=0 ⟹m=2,−3$

Hence,

$$C.F=ϕ_ 1 ​ (y+2x)+ϕ _2 ​ (y−3x)$$

Now,

$$P.I=\frac{1}{(D ^ 2 +DD ^′ −6D ^{′2} )}ycosx$$

$\implies P.I=\frac{1}{(D+3D^{'})(D-2D^{'})}ycosx=\frac{1}{D+3D^{'}}\int_{D-2D^{'}}ycosxdx=\frac{1}{D+3D^{'}}[(c_1-2x)sinx-2cosx]=\frac{1}{D+3D^{'}}(ysinx-2cosx)=\int_{D+3D^{'}}(ysinxdx)-\int_{D+3D^{'}}2cosxdx=\int _{D+3D^{'}}(c_2+3x)sinxdx=\int _{D+3D^{'}}2cosxdx$

$$(y=c_2+3x)$$

$$\implies P.I=-ycosx+\int cosxdx = -ycosx +sinx$$

Therefore the complete solution is

$C.F+P.I= \phi_1(y_+2x)+\phi_2(y-3x)+-ycosx+sinx$