solve the PDE (D^2+DD'-6D'^2)z = x^2sin(x+y)
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Answer to Question #139807 in Differential Equations for Nikhil Singh
Answers>Math>Differential Equations
Question #139807
Solve
(D^2+DD'-6D'^2)z= ycosx
Expert's answer
Given,
(D^2+DD'-6D'^2)z= y\cos x(D
2
+DD
′
−6D
′2
)z=ycosx
Now, Auxiliary equation will be
m^2+m-6=0\\ \implies (m-2)(m+3)=0\\ \implies m=2,-3m
2
+m−6=0
⟹(m−2)(m+3)=0
⟹m=2,−3
Hence,
C.F=\phi_1(y+2x)+\phi_2(y-3x)C.F=ϕ
1
(y+2x)+ϕ
2
(y−3x)
Now,
P.I=\frac{1}{(D^2+DD'-6D'^2)}y\cos x\\ \implies P.I=\frac{1}{(D+3D')(D-2D')}y\cos x\\ =\frac{1}{D+3D'}\int_{D-2D'}y\cos x \, dx\\ =\frac{1}{D+3D'}[(c_1-2x)\sin x-2\cos x]\\ =\frac{1}{D+3D'}(y\sin x-2\cos x)\\ =\int_{D+3D'}y\sin xdx-\int_{D+3D'}2\cos x dx\\ =\int_{D+3D'}(c_2+3x)\sin xdx-\int_{D+3D'}2\cos x dx\hspace{1cm}(y=c_2+3x)\\P.I=
(D
2
+DD
′
−6D
′2
)
1
ycosx
⟹P.I=
(D+3D
′
)(D−2D
′
)
1
ycosx
=
D+3D
′
1
∫
D−2D
′
ycosxdx
=
D+3D
′
1
[(c
1
−2x)sinx−2cosx]
=
D+3D
′
1
(ysinx−2cosx)
=∫
D+3D
′
ysinxdx−∫
D+3D
′
2cosxdx
=∫
D+3D
′
(c
2
+3x)sinxdx−∫
D+3D
′
2cosxdx(y=c
2
+3x)
\implies P.I=-y\cos x+\int\cos x \, dx =-y\cos x+\sin x⟹P.I=−ycosx+∫cosxdx=−ycosx+sinx
Therefore the complete solution is
C.F+P.I=\phi_1(y+2x)+\phi_2(y-3x)+-y\cos x+\sin xC.F+P.I=ϕ
1
(y+2x)+ϕ
2
(y−3x)+−ycosx+sinx
Step-by-step explanation:
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