Answer:-
:-I will assume that on the right side of the equation it looks like this
( D 2 − 3 D D ′ + 2 D ′ 2 ) z = e 2 x − y + e x + y \left(D^2-3DD'+2D'^2\right)z=e^{2x-y}+e^{x+y} ( D 2 − 3 D D ′ + 2 D ′2 ) z = e 2 x − y + e x + y This is an inhomogeneous equation, so its solution consists of two parts
z ( x , y ) = ( C . F . ) + ( P . I . ) ( C . F . ) − Complementary Function of Homogeneous Linear PDE ( P . I . ) − Particular Integral z(x,y)=(C.F.)+(P.I.)\\[0.3cm]
(C.F.)-\text{Complementary Function of Homogeneous Linear PDE}\\[0.3cm]
(P.I.)-\text{Particular Integral} z ( x , y ) = ( C . F . ) + ( P . I . ) ( C . F . ) − Complementary Function of Homogeneous Linear PDE ( P . I . ) − Particular Integral 1 STEP : We try to find ( C . F . ) (C.F.) ( C . F . )
As we know, the equation
( D − m D ′ ) z = 0 → z ( x , y ) = ϕ ( y + m x ) ϕ ( y + m x ) is an arbitrary function \boxed{\left(D-mD'\right)z=0\to z(x,y)=\phi(y+mx)}\\[0.3cm]
\phi(y+mx)\,\text{is an arbitrary function} ( D − m D ′ ) z = 0 → z ( x , y ) = ϕ ( y + m x ) ϕ ( y + m x ) is an arbitrary function In our case,
( D 2 − 3 D D ′ + 2 D ′ 2 ) z = ( D 2 − D D ′ − 2 D D ′ + 2 D ′ 2 ) z = = ( D ( D − D ′ ) − 2 D ′ ( D − D ′ ) ) z = = ( D − D ′ ) ( D − 2 D ′ ) z ( D − D ′ ) ( D − 2 D ′ ) z = 0 ⟶ { ( D − 1 D ′ ) z = 0 → z ( x , y ) = ϕ 1 ( y + x ) ( D − 2 D ′ ) z = 0 → z ( x , y ) = ϕ 2 ( y + 2 x ) \left(D^2-3DD'+2D'^2\right)z=\left(D^2-DD'-2DD'+2D'^2\right)z=\\[0.3cm]
=\left(D\left(D-D'\right)-2D'\left(D-D'\right)\right)z=\\[0.3cm]
=\left(D-D'\right)\left(D-2D'\right)z\\[0.3cm]
\left(D-D'\right)\left(D-2D'\right)z=0\longrightarrow\\[0.3cm]
\left\{\begin{array}{l}
\left(D-1D'\right)z=0\to z(x,y)=\phi_1(y+x)\\[0.3cm]
\left(D-2D'\right)z=0\to z(x,y)=\phi_2(y+2x)
\end{array}\right. ( D 2 − 3 D D ′ + 2 D ′2 ) z = ( D 2 − D D ′ − 2 D D ′ + 2 D ′2 ) z = = ( D ( D − D ′ ) − 2 D ′ ( D − D ′ ) ) z = = ( D − D ′ ) ( D − 2 D ′ ) z ( D − D ′ ) ( D − 2 D ′ ) z = 0 ⟶ { ( D − 1 D ′ ) z = 0 → z ( x , y ) = ϕ 1 ( y + x ) ( D − 2 D ′ ) z = 0 → z ( x , y ) = ϕ 2 ( y + 2 x ) Conclusion,
( C . F . ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) ϕ 1 , ϕ 2 are arbitrary functions \boxed{(C.F.)=\phi_1(y+x)+\phi_2(y+2x)}\\[0.3cm]
\phi_1,\phi_2\,\text{are arbitrary functions} ( C . F . ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) ϕ 1 , ϕ 2 are arbitrary functions 2 STEP : We try to find ( P . I . ) (P.I.) ( P . I . )
As we know
f ( D , D ′ ) z = F ( x , y ) → ( P . I . ) = 1 f ( D , D ′ ) F ( x , y ) special case : 1 f ( D , D ′ ) e a x + b y = e a x + b y f ( a , b ) , f ( a , b ) ≠ 0 i f f ( a , b ) = 0 : ( P . I . ) = x ⋅ 1 f ′ ( D , D ′ ) ⋅ e a x + b y , f ′ ( D , D ′ ) is the partial derivative of f ( D , D ′ ) w.r.t D f\left(D,D'\right)z=F(x,y)\to(P.I.)=\frac{1}{f\left(D,D'\right)}F(x,y)\\[0.3cm]
\text{special case :}\,\,\,\frac{1}{f\left(D,D'\right)}e^{ax+by}=\frac{e^{ax+by}}{f(a,b)},\,\,\,f(a,b)\neq0\\[0.3cm]
if\,\,\,f(a,b)=0 : (P.I.)=x\cdot\frac{1}{f'\left(D,D'\right)}\cdot e^{ax+by},\\[0.3cm]
f'\left(D,D'\right)\text{ is the partial derivative of}\,\,\,f\left(D,D'\right)\,\,\,\text{ w.r.t D} f ( D , D ′ ) z = F ( x , y ) → ( P . I . ) = f ( D , D ′ ) 1 F ( x , y ) special case : f ( D , D ′ ) 1 e a x + b y = f ( a , b ) e a x + b y , f ( a , b ) = 0 i f f ( a , b ) = 0 : ( P . I . ) = x ⋅ f ′ ( D , D ′ ) 1 ⋅ e a x + b y , f ′ ( D , D ′ ) is the partial derivative of f ( D , D ′ ) w.r.t D In our case,
( D 2 − 3 D D ′ + 2 D ′ 2 ) z = e 2 x − y ⏟ ( P . I . ) 1 + e x + y ⏟ ( P . I . ) 2 ( P . I . ) 1 : ( D 2 − 3 D D ′ + 2 D ′ 2 ) z = e 2 x − y → ( P . I . ) 1 = 1 ( D 2 − 3 D D ′ + 2 D ′ 2 ) ⋅ e 2 x − y = e 2 x − y 2 2 − 3 ⋅ 2 ⋅ ( − 1 ) + 2 ( − 1 ) 2 = = e 2 x − y 4 + 6 + 2 = e 2 x − y 12 → ( P . I . ) 1 = e 2 x − y 12 ( P . I . ) 2 = 1 ( D 2 − 3 D D ′ + 2 D ′ 2 ) ⋅ e x + y = x ⋅ 1 2 D − 3 D ′ ⋅ e x + y = = x ⋅ e x + y 2 − 3 = − x ⋅ e x + y → ( P . I . ) 2 = − x ⋅ e x + y \left(D^2-3DD'+2D'^2\right)z=\underbrace{e^{2x-y}}_{(P.I.)_1}+\underbrace{e^{x+y}}_{(P.I.)_2}\\[0.3cm]
(P.I.)_1 : \left(D^2-3DD'+2D'^2\right)z=e^{2x-y}\to\\[0.3cm]
(P.I.)_1=\frac{1}{\left(D^2-3DD'+2D'^2\right)}\cdot e^{2x-y}=\frac{e^{2x-y}}{2^2-3\cdot2\cdot(-1)+2(-1)^2}=\\[0.3cm]
=\frac{e^{2x-y}}{4+6+2}=\frac{e^{2x-y}}{12}\to\boxed{(P.I.)_1=\frac{e^{2x-y}}{12}}\\[0.3cm]
(P.I.)_2=\frac{1}{\left(D^2-3DD'+2D'^2\right)}\cdot e^{x+y}=x\cdot\frac{1}{2D-3D'}\cdot e^{x+y}=\\[0.3cm]
=\frac{x\cdot e^{x+y}}{2-3}=-x\cdot e^{x+y}\to\boxed{(P.I.)_2=-x\cdot e^{x+y}} ( D 2 − 3 D D ′ + 2 D ′2 ) z = ( P . I . ) 1 e 2 x − y + ( P . I . ) 2 e x + y ( P . I . ) 1 : ( D 2 − 3 D D ′ + 2 D ′2 ) z = e 2 x − y → ( P . I . ) 1 = ( D 2 − 3 D D ′ + 2 D ′2 ) 1 ⋅ e 2 x − y = 2 2 − 3 ⋅ 2 ⋅ ( − 1 ) + 2 ( − 1 ) 2 e 2 x − y = = 4 + 6 + 2 e 2 x − y = 12 e 2 x − y → ( P . I . ) 1 = 12 e 2 x − y ( P . I . ) 2 = ( D 2 − 3 D D ′ + 2 D ′2 ) 1 ⋅ e x + y = x ⋅ 2 D − 3 D ′ 1 ⋅ e x + y = = 2 − 3 x ⋅ e x + y = − x ⋅ e x + y → ( P . I . ) 2 = − x ⋅ e x + y General conclusion.
( D 2 − 3 D D ′ + 2 D ′ 2 ) z = e 2 x − y + e x + y → z ( x , y ) = ( C . F . ) + ( P . I . ) 1 + ( P . I . ) 2 z ( x , y ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) + e 2 x − y 12 − x ⋅ e x + y ϕ 1 , ϕ 2 are arbitrary functions \left(D^2-3DD'+2D'^2\right)z=e^{2x-y}+e^{x+y}\to\\[0.3cm]
z(x,y)=(C.F.)+(P.I.)_1+(P.I.)_2\\[0.3cm]
\boxed{z(x,y)=\phi_1(y+x)+\phi_2(y+2x)+\frac{e^{2x-y}}{12}-x\cdot e^{x+y}}\\[0.3cm]
\phi_1,\phi_2\,\text{are arbitrary functions} ( D 2 − 3 D D ′ + 2 D ′2 ) z = e 2 x − y + e x + y → z ( x , y ) = ( C . F . ) + ( P . I . ) 1 + ( P . I . ) 2 z ( x , y ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) + 12 e 2 x − y − x ⋅ e x + y ϕ 1 , ϕ 2 are arbitrary functions ANSWER
z ( x , y ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) + e 2 x − y 12 − x ⋅ e x + y ϕ 1 , ϕ 2 are arbitrary functions z(x,y)=\phi_1(y+x)+\phi_2(y+2x)+\frac{e^{2x-y}}{12}-x\cdot e^{x+y}\\[0.3cm]
\phi_1,\phi_2\,\text{are arbitrary functions} z ( x , y ) = ϕ 1 ( y + x ) + ϕ 2 ( y + 2 x ) + 12 e 2 x − y − x ⋅ e x + y ϕ 1 , ϕ 2 are arbitrary functions 
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