Question

Solve
(D^3-DD'^2+D^2+DD')z=(x+2)+e^(x-y)

Answer 1

The given partial differential equation is

$$(D^3-DD'^2+D^2+DD')z=(x+2)+e^{x-y}$$

This is non homogeneous linear partial differential equation with constant coefficients .The solution of the given PDE is obtained by finding C.F and P.I

Thus ,solution = C.F + P.I

Calculating C.F

$D^3-DD'^2+D^2+DD'=0\\or, D(D^2-D'^2+D+D')=0\\or, D\{(D+D')(D-D')+(D+D')\}=0\\or, D(D+D')(D-D'+1)=0$

Now comparing with $(D-m_1D'-c_1)(D-m_2D'-c_2)(D-m_3D'-c_3)z=0,we \space get$

$Rewriting \space D(D+D')(D-D'+1)=0 \space as\\(D-0*D'-0)(D-(-1)*D'-0)(D-1*D'+1)=0$

$\therefore m_1=0,m_2=-1,m_3=1\\and \space c_1=0,c_2=0,c_3=1$

$\therefore C.F = e^{0*x}f_1(y+0*x)+ e^{0*x}f_2(y-x)+e^{1*x}f_3(y+x)\\or, C.F = f_1(y+0*x)+ f_2(y-x)+e^{x}f_3(y+x)\\C.F = e^{0*x}f_1(y+0*x)+ e^{0*x}f_2(y-x)+e^{1*x}f_3(y+x)\\or, C.F = f_1(y)+ f_2(y-x)+e^{x}f_3(y+x)$

Calculating P.I

$P.I = \frac{1}{f(D,D')}\{e^{x-y}+x+2\}\\or, P.I = \frac{1}{f(D,D')}e^{x-y}+\frac{1}{f(D,D')}(x+2)\\or,P.I = (P.I)_1+(P.I)_2$

$Now, (P.I)_1 = \frac{1}{f(D,D')}e^{x-y}\\$

$= \frac{1}{D^3-DD'^2+D^2+DD'}e^{x-y}$

Now replacing D by 1 and D' by (-1) we get $D^3-DD'^2+D^2+DD' = 0,$ so following the working rule

$(P.I)_1 = \frac{x}{\frac{\delta}{\delta D}D^3-DD'^2+D^2+DD'} e^{x-y}\\$

$= \frac{x}{3D^2-D'^2+2D+D'} e^{x-y}$

$= \frac{x}{3-1+2-1} e^{x-y}$

$=\frac{x}{3}e^{x-y}$

$(P.I)_1 = \frac{x}{3}e^{x-y}.$

$Now, (P.I)_2 = \frac{1}{f(D,D')}(x+2)$

$= \frac{1}{ D(D+D')(D-D'+1)} (x+2)$

$=\frac{1}{D(D+D')}\{1+(D-D')\}^{-1} \space (x+2)$

$=\frac{1}{D(D+D')}(1+(D-D')+(D-D')^2......) \space (x+2)$

$= x +2+1$

$=x+3$ $\because D(x)=1 \space D'(1)=0$

$(P.I)_2 = x+3$

$\therefore solution \space = C.F+P.I\\$

$= f_1(y)+ f_2(y-x)+e^{x}f_3(y+x) \space + \frac{x}{3}e^{x-y} + (x+3)$

ANSWER