東南電機系大學部二技 工程數學作業(5)
01) Solve y′′ +3y′ +2y=ex.
x x
x 3
+ e3
1
h
02) Solve y′′ −2y′ −3y e= −x. 03) Solve y′ −2y=cos3 . 04) Solve + 3y′′ y′ −4y=sin2 . 05) Solve +y′′ 4y=sin2x+cos . 06) Solve
c
D2−3D+ 2h
y e x= xb g
1 . 07) Solvec
D2−4h
y= x2 x.08) Solve
c
D2−4D+3h
y=x2 + +x . 09) Solvec
D3−2D2−5D+6h c
y= e2x+3 2.10) Solve D D y .
ex
2 1
+ = 1
c h
+ 11) To solve yu D a x bx c
p = 1 2+
b g c
+h
by using long-division process, we propose a new idea to expand the operator 1u D
b g
as Taylor’s series expansion with respect to D, that is,
1 0 0 0
2
0 1
0 0 2
2
0
2 0
2
3
0
u D f D f f D f
D
f u f u
u f u u u
D D u D
b g b g
≡ = + ′ ⋅ + ′′ ⋅ +
= ′ = − ′ ′′ = ′ − ⋅ ′′
= = =
( ) ( ) ( ) ( )
!
( ) , ( ) , ( )
LL
where
a) Proves above statement first.
b) Using the result above to find out the particular solution of ,
that is,
2D2+2D+3 yp = x2+2x−
c h
1yp = u D1 x + x
2 1
2
( )