Answer to Question #229305 in Differential Equations for Hridoy

Question

Solve it
(D^2+DD'-6D'^2)z= ysinx

Answer 1

Auxiliary equation will be

$$m^2+m-6=0$$

$$(m-2)(m+3)=0$$

$$m_1=2, m_2=-3$$

Hence

$$C.F.=\phi_1(y+2x)+\phi_2(y-3x)$$

$$P.I.=\dfrac{1}{D^2+DD'-6D'^2}y\sin x$$

$$=I.P.\ of\ e^{ix}\dfrac{1}{(D+i)^2+(D+i)(D'+0)-6(D'-0)^2}y$$

$$=I.P.\ of\ e^{ix}\dfrac{1}{D^2-1+2iD+DD'+iD'-6D'^2}y$$

$$=I.P.\ of\ e^{ix}\dfrac{1}{D^2+DD'-6D'^2+i(2D+D')-1}y$$

$$=I.P.\ of\ -e^{ix}\dfrac{1}{1-[D^2+DD'-6D'^2+i(2D+D')]}y$$

Using binomial expansion

$$P.I.=I.P.\ of\ -e^{ix}(1+[i(2D+D')+D^2+DD'-6D'^2])y$$

$$=I.P.\ of\ -e^{ix}(y+i)$$

$$=I.P.\ of\ -(\cos x-i\sin x)(y+i)$$

$$P.I.=-\cos x-y\sin x$$

Hence the general solution is given by

$$y=\phi_1(y+2x)+\phi_2(y-3x)-(\cos x+y\sin x)$$