The P.I. of (D^3 – 2D²D') 2 = e^x+2y
Answers
Answer:
Step-by-step explanation:
OLUTION
Let us use some known facts. Let the given differential equation be
F(D,D')=f(x,y)F(D,D
′
)=f(x,y)
Factorize F(D,D')F(D,D
′
) into linear factors. Then use the following results:
Corresponding to each non-repeated factor \left(bD-aD'-c\right)(bD−aD
′
−c) , the part of C.F. is taken as
\exp\left(\frac{cx}{b}\right)\cdot\varphi(by+ax),\quad if\quad b\neq0exp(
b
cx
)⋅φ(by+ax),ifb
=0
In our case,
\left(D^2+3DD'+2D'^2\right)z=x+y\longrightarrow z(x,y)=C.F.+P.I.(D
2
+3DD
′
+2D
′2
)z=x+y⟶z(x,y)=C.F.+P.I.
0 STEP: We factor the expression
\left(D^2+3DD'+2D'^2\right)=\left(D^2+DD'\right)+\left(2DD'+2D'^2\right)=(D
2
+3DD
′
+2D
′2
)=(D
2
+DD
′
)+(2DD
′
+2D
′2
)=
=D\left(D+D'\right)+2D'\left(D+D'\right)=\left(D+D'\right)\left(D+2D'\right)=D(D+D
′
)+2D
′
(D+D
′
)=(D+D
′
)(D+2D
′
)
Conclusion,
\boxed{\left(D^2+3DD'+2D'^2\right)z=\left(D+D'\right)\left(D+2D'\right)z}
(D
2
+3DD
′
+2D
′2
)z=(D+D
′
)(D+2D
′
)z
1 STEP: Let find C.F.