Aptarimas : Matematika/Antrosios eilės tiesinės nehomogeninės diferencialinės lygtys
y
″
−
5
y
′
+
6
y
=
f
(
x
)
,
{\displaystyle y''-5y'+6y=f(x),}
kai: a)
f
(
x
)
=
2
x
2
e
x
;
{\displaystyle f(x)=2x^{2}e^{x};}
f
(
x
)
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle f(x)=(2x-3)e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
5
y
′
+
6
y
=
0.
{\displaystyle y''-5y'+6y=0.}
k
2
−
5
k
+
6
=
0
{\displaystyle k^{2}-5k+6=0}
k
1
=
2
{\displaystyle k_{1}=2}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
″
−
5
y
′
+
6
y
=
0
{\displaystyle y''-5y'+6y=0}
y
¯
=
C
1
e
2
x
+
C
2
e
3
x
.
{\displaystyle {\bar {y}}=C_{1}e^{2x}+C_{2}e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
a) Kai
f
(
x
)
=
2
x
2
e
x
,
{\displaystyle f(x)=2x^{2}e^{x},}
P
n
(
x
)
=
2
x
2
{\displaystyle P_{n}(x)=2x^{2}}
α
=
1.
{\displaystyle \alpha =1.}
α
=
1
{\displaystyle \alpha =1}
y
~
=
(
a
x
2
+
b
x
+
c
)
e
x
{\displaystyle {\tilde {y}}=(ax^{2}+bx+c)e^{x}}
(daugianaris
Q
n
(
x
)
=
a
x
2
+
b
x
+
c
,
{\displaystyle Q_{n}(x)=ax^{2}+bx+c,}
a
x
+
b
,
{\displaystyle ax+b,}
a
x
3
+
b
x
2
+
c
x
+
d
{\displaystyle ax^{3}+bx^{2}+cx+d}
Toliau randame:
y
~
′
=
(
(
a
x
2
+
b
x
+
c
)
e
x
)
′
=
2
a
x
e
x
+
a
x
2
e
x
+
b
e
x
+
b
x
e
x
+
c
e
x
=
{\displaystyle {\tilde {y}}'=((ax^{2}+bx+c)e^{x})'=2axe^{x}+ax^{2}e^{x}+be^{x}+bxe^{x}+ce^{x}=}
=
(
a
x
2
+
2
a
x
+
b
x
+
b
+
c
)
e
x
,
{\displaystyle =(ax^{2}+2ax+bx+b+c)e^{x},}
y
~
″
=
(
a
x
2
+
4
a
x
+
b
x
+
2
a
+
2
b
+
c
)
e
x
.
{\displaystyle {\tilde {y}}''=(ax^{2}+4ax+bx+2a+2b+c)e^{x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
a
x
2
+
4
a
x
+
b
x
+
2
a
+
2
b
+
c
)
e
x
−
5
(
a
x
2
+
2
a
x
+
b
x
+
b
+
c
)
e
x
+
6
(
a
x
2
+
b
x
+
c
)
e
x
=
2
x
2
e
x
,
{\displaystyle (ax^{2}+4ax+bx+2a+2b+c)e^{x}-5(ax^{2}+2ax+bx+b+c)e^{x}+6(ax^{2}+bx+c)e^{x}=2x^{2}e^{x},}
(
a
x
2
+
4
a
x
+
b
x
+
2
a
+
2
b
+
c
)
−
5
(
a
x
2
+
2
a
x
+
b
x
+
b
+
c
)
+
6
(
a
x
2
+
b
x
+
c
)
=
2
x
2
,
{\displaystyle (ax^{2}+4ax+bx+2a+2b+c)-5(ax^{2}+2ax+bx+b+c)+6(ax^{2}+bx+c)=2x^{2},}
a
x
2
+
4
a
x
+
b
x
+
2
a
+
2
b
+
c
−
5
a
x
2
−
10
a
x
−
5
b
x
−
5
b
−
5
c
+
6
a
x
2
+
6
b
x
+
6
c
=
2
x
2
,
{\displaystyle ax^{2}+4ax+bx+2a+2b+c-5ax^{2}-10ax-5bx-5b-5c+6ax^{2}+6bx+6c=2x^{2},}
2
a
x
2
−
6
a
x
+
2
b
x
+
2
a
−
3
b
+
2
c
=
2
x
2
.
{\displaystyle 2ax^{2}-6ax+2bx+2a-3b+2c=2x^{2}.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
2
|
2
a
=
2
,
{\displaystyle x^{2}\quad |\;2a=2,}
x
1
|
−
6
a
+
2
b
=
0
,
|
|
(
2
b
=
6
,
b
=
3
)
{\displaystyle x^{1}\quad |\;-6a+2b=0,\quad ||\;(2b=6,\;b=3)}
x
0
|
2
a
−
3
b
+
2
c
=
0.
|
|
(
2
−
9
+
2
c
=
0
,
c
=
−
7
/
2
=
3.5
)
{\displaystyle x^{0}\quad |\;2a-3b+2c=0.\quad ||\;(2-9+2c=0,\;c=-7/2=3.5)}
Iš jos randame:
a
=
1
,
b
=
3
,
c
=
3.5.
{\displaystyle a=1,\;b=3,\;c=3.5.}
y
~
=
(
x
2
+
3
x
+
3.5
)
e
x
{\displaystyle {\tilde {y}}=(x^{2}+3x+3.5)e^{x}}
y
=
y
¯
+
y
~
=
C
1
e
2
x
+
C
2
e
3
x
+
(
x
2
+
3
x
+
3.5
)
e
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{2x}+C_{2}e^{3x}+(x^{2}+3x+3.5)e^{x}.}
b) Kai
f
(
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle f(x)=(2x-3)e^{3x},}
P
n
(
x
)
=
2
x
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=2x-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
x
+
b
;
α
=
3
{\displaystyle Q_{n}(x)=ax+b;\;\alpha =3}
y
~
=
x
(
a
x
+
b
)
e
3
x
=
(
a
x
2
+
b
x
)
e
3
x
.
{\displaystyle {\tilde {y}}=x(ax+b)e^{3x}=(ax^{2}+bx)e^{3x}.}
Toliau sprendžiama analogiškai a) atvejui.
x
2
|
2
a
=
0
,
|
|
(
a
=
0
/
2
=
0
)
{\displaystyle x^{2}\quad |\;2a=0,\quad ||\;(a=0/2=0)}
x
1
|
−
6
a
+
2
b
=
2
,
|
|
(
2
b
=
2
,
b
=
1
)
{\displaystyle x^{1}\quad |\;-6a+2b=2,\quad ||\;(2b=2,\;b=1)}
x
0
|
2
a
−
3
b
+
2
c
=
−
3.
|
|
(
2
⋅
0
−
3
⋅
1
+
2
c
=
−
3
,
2
c
=
0
,
c
=
0
)
{\displaystyle x^{0}\quad |\;2a-3b+2c=-3.\quad ||\;(2\cdot 0-3\cdot 1+2c=-3,\;2c=0,\;c=0)}
Iš sistemos randame:
a
=
0
,
b
=
1.
{\displaystyle a=0,\;b=1.}
y
~
=
(
0
⋅
x
2
+
1
⋅
x
)
e
3
x
=
x
e
3
x
.
{\displaystyle {\tilde {y}}=(0\cdot x^{2}+1\cdot x)e^{3x}=xe^{3x}.}
y
″
−
5
y
′
+
6
y
=
(
2
x
−
3
)
e
3
x
{\displaystyle y''-5y'+6y=(2x-3)e^{3x}}
y
=
y
¯
+
y
~
=
C
1
e
2
x
+
C
2
e
3
x
+
x
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{2x}+C_{2}e^{3x}+xe^{3x}.}
Patikriname:
y
~
′
=
(
x
e
3
x
)
′
=
e
3
x
+
3
x
e
3
x
;
{\displaystyle {\tilde {y}}'=(xe^{3x})'=e^{3x}+3xe^{3x};}
y
~
″
=
(
e
3
x
+
3
x
e
3
x
)
′
=
3
e
3
x
+
3
e
3
x
+
9
x
e
3
x
=
6
e
3
x
+
9
x
e
3
x
;
{\displaystyle {\tilde {y}}''=(e^{3x}+3xe^{3x})'=3e^{3x}+3e^{3x}+9xe^{3x}=6e^{3x}+9xe^{3x};}
y
″
−
5
y
′
+
6
y
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle y''-5y'+6y=(2x-3)e^{3x},}
6
e
3
x
+
9
x
e
3
x
−
5
(
e
3
x
+
3
x
e
3
x
)
+
6
x
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle 6e^{3x}+9xe^{3x}-5(e^{3x}+3xe^{3x})+6xe^{3x}=(2x-3)e^{3x},}
6
e
3
x
+
9
x
e
3
x
−
5
e
3
x
−
15
x
e
3
x
+
6
x
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle 6e^{3x}+9xe^{3x}-5e^{3x}-15xe^{3x}+6xe^{3x}=(2x-3)e^{3x},}
e
3
x
+
9
x
e
3
x
−
9
x
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle e^{3x}+9xe^{3x}-9xe^{3x}=(2x-3)e^{3x},}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
6
y
′
+
9
y
=
0.
{\displaystyle y''-6y'+9y=0.}
k
2
−
6
k
+
9
=
0
{\displaystyle k^{2}-6k+9=0}
k
1
=
3
{\displaystyle k_{1}=3}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
¯
=
(
C
1
+
x
C
2
)
e
3
x
.
{\displaystyle {\bar {y}}=(C_{1}+xC_{2})e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
Kai
f
(
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle f(x)=(2x-3)e^{3x},}
P
n
(
x
)
=
2
x
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=2x-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
x
+
b
;
α
=
3
{\displaystyle Q_{n}(x)=ax+b;\;\alpha =3}
y
~
=
x
2
(
a
x
+
b
)
e
3
x
=
(
a
x
3
+
b
x
2
)
e
3
x
.
{\displaystyle {\tilde {y}}=x^{2}(ax+b)e^{3x}=(ax^{3}+bx^{2})e^{3x}.}
Randame išvestines:
y
~
′
=
(
(
a
x
3
+
b
x
2
)
e
3
x
)
′
=
3
a
x
2
e
3
x
+
3
a
x
3
e
3
x
+
2
b
x
e
3
x
+
3
b
x
2
e
3
x
=
{\displaystyle {\tilde {y}}'=((ax^{3}+bx^{2})e^{3x})'=3ax^{2}e^{3x}+3ax^{3}e^{3x}+2bxe^{3x}+3bx^{2}e^{3x}=}
=
(
3
a
x
3
+
3
a
x
2
+
3
b
x
2
+
2
b
x
)
e
3
x
,
{\displaystyle =(3ax^{3}+3ax^{2}+3bx^{2}+2bx)e^{3x},}
y
~
″
=
(
9
a
x
2
+
6
a
x
+
6
b
x
+
2
b
)
e
3
x
+
3
(
3
a
x
3
+
3
a
x
2
+
3
b
x
2
+
2
b
x
)
e
3
x
=
{\displaystyle {\tilde {y}}''=(9ax^{2}+6ax+6bx+2b)e^{3x}+3(3ax^{3}+3ax^{2}+3bx^{2}+2bx)e^{3x}=}
=
(
9
a
x
3
+
18
a
x
2
+
6
a
x
+
9
b
x
2
+
12
b
x
+
2
b
)
e
3
x
.
{\displaystyle =(9ax^{3}+18ax^{2}+6ax+9bx^{2}+12bx+2b)e^{3x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
9
a
x
3
+
18
a
x
2
+
6
a
x
+
9
b
x
2
+
12
b
x
+
2
b
)
e
3
x
−
6
(
3
a
x
3
+
3
a
x
2
+
3
b
x
2
+
2
b
x
)
e
3
x
+
9
(
a
x
2
+
b
x
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle (9ax^{3}+18ax^{2}+6ax+9bx^{2}+12bx+2b)e^{3x}-6(3ax^{3}+3ax^{2}+3bx^{2}+2bx)e^{3x}+9(ax^{2}+bx)e^{3x}=(2x-3)e^{3x},}
9
a
x
3
+
18
a
x
2
+
6
a
x
+
9
b
x
2
+
12
b
x
+
2
b
−
18
a
x
3
−
18
a
x
2
−
18
b
x
2
−
12
b
x
+
9
a
x
2
+
9
b
x
=
2
x
−
3
,
{\displaystyle 9ax^{3}+18ax^{2}+6ax+9bx^{2}+12bx+2b-18ax^{3}-18ax^{2}-18bx^{2}-12bx+9ax^{2}+9bx=2x-3,}
9
a
x
3
−
18
a
x
3
+
18
a
x
2
−
18
a
x
2
+
9
a
x
2
+
6
a
x
+
9
b
x
2
−
18
b
x
2
+
12
b
x
−
12
b
x
+
9
b
x
+
2
b
=
2
x
−
3
,
{\displaystyle 9ax^{3}-18ax^{3}+18ax^{2}-18ax^{2}+9ax^{2}+6ax+9bx^{2}-18bx^{2}+12bx-12bx+9bx+2b=2x-3,}
−
9
a
x
3
+
9
a
x
2
+
6
a
x
−
9
b
x
2
+
9
b
x
+
2
b
=
2
x
−
3.
{\displaystyle -9ax^{3}+9ax^{2}+6ax-9bx^{2}+9bx+2b=2x-3.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
3
|
−
9
a
=
0
,
{\displaystyle x^{3}\quad |\;-9a=0,}
x
2
|
9
a
−
9
b
=
0
,
{\displaystyle x^{2}\quad |\;9a-9b=0,}
x
1
|
6
a
+
9
b
=
2
,
{\displaystyle x^{1}\quad |\;6a+9b=2,}
x
0
|
2
b
=
−
3.
{\displaystyle x^{0}\quad |\;2b=-3.}
Iš sistemos randame:
a
=
(
2
+
9
⋅
1.5
)
/
6
=
(
2
+
13.5
)
/
6
=
2.58
(
3
)
=
31
/
12
,
b
=
−
3
/
2
=
−
1.5.
{\displaystyle a=(2+9\cdot 1.5)/6=(2+13.5)/6=2.58(3)=31/12,\;b=-3/2=-1.5.}
y
~
=
(
31
12
x
2
−
3
2
x
)
e
3
x
.
{\displaystyle {\tilde {y}}=({\frac {31}{12}}x^{2}-{\frac {3}{2}}x)e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}}
y
=
y
¯
+
y
~
=
C
1
e
3
x
+
x
C
2
e
3
x
+
(
31
12
x
2
−
3
2
x
)
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{3x}+xC_{2}e^{3x}+({\frac {31}{12}}x^{2}-{\frac {3}{2}}x)e^{3x}.}
Patikriname:
y
~
′
=
(
(
31
12
x
2
−
3
2
x
)
e
3
x
)
′
=
(
62
12
x
−
3
2
)
e
3
x
+
3
(
31
12
x
2
−
3
2
x
)
e
3
x
=
{\displaystyle {\tilde {y}}'=(({\frac {31}{12}}x^{2}-{\frac {3}{2}}x)e^{3x})'=({\frac {62}{12}}x-{\frac {3}{2}})e^{3x}+3({\frac {31}{12}}x^{2}-{\frac {3}{2}}x)e^{3x}=}
=
(
31
12
x
2
+
62
12
x
−
9
2
x
−
3
2
)
e
3
x
=
(
31
12
x
2
+
62
−
9
⋅
6
12
x
−
3
2
)
e
3
x
=
{\displaystyle =({\frac {31}{12}}x^{2}+{\frac {62}{12}}x-{\frac {9}{2}}x-{\frac {3}{2}})e^{3x}=({\frac {31}{12}}x^{2}+{\frac {62-9\cdot 6}{12}}x-{\frac {3}{2}})e^{3x}=}
=
(
31
12
x
2
+
8
12
x
−
3
2
)
e
3
x
=
(
31
12
x
2
+
2
3
x
−
3
2
)
e
3
x
,
{\displaystyle =({\frac {31}{12}}x^{2}+{\frac {8}{12}}x-{\frac {3}{2}})e^{3x}=({\frac {31}{12}}x^{2}+{\frac {2}{3}}x-{\frac {3}{2}})e^{3x},}
y
~
″
=
(
(
31
12
x
2
+
2
3
x
−
3
2
)
e
3
x
)
′
=
(
62
12
x
+
2
3
)
e
3
x
+
3
(
31
12
x
2
+
2
3
x
−
3
2
)
e
3
x
=
{\displaystyle {\tilde {y}}''=(({\frac {31}{12}}x^{2}+{\frac {2}{3}}x-{\frac {3}{2}})e^{3x})'=({\frac {62}{12}}x+{\frac {2}{3}})e^{3x}+3({\frac {31}{12}}x^{2}+{\frac {2}{3}}x-{\frac {3}{2}})e^{3x}=}
=
31
6
x
e
3
x
+
2
3
e
3
x
+
93
12
x
2
e
3
x
+
6
3
x
e
3
x
−
9
2
e
3
x
=
{\displaystyle ={\frac {31}{6}}xe^{3x}+{\frac {2}{3}}e^{3x}+{\frac {93}{12}}x^{2}e^{3x}+{\frac {6}{3}}xe^{3x}-{\frac {9}{2}}e^{3x}=}
=
(
93
12
x
2
+
31
6
x
+
1
2
x
+
2
3
−
9
2
)
e
3
x
=
(
93
12
x
2
+
62
−
6
12
x
+
4
−
27
6
)
e
3
x
=
{\displaystyle =({\frac {93}{12}}x^{2}+{\frac {31}{6}}x+{\frac {1}{2}}x+{\frac {2}{3}}-{\frac {9}{2}})e^{3x}=({\frac {93}{12}}x^{2}+{\frac {62-6}{12}}x+{\frac {4-27}{6}})e^{3x}=}
=
(
93
12
x
2
+
54
12
x
−
23
6
)
e
3
x
=
(
31
4
x
2
+
9
2
x
−
23
6
)
e
3
x
;
{\displaystyle =({\frac {93}{12}}x^{2}+{\frac {54}{12}}x-{\frac {23}{6}})e^{3x}=({\frac {31}{4}}x^{2}+{\frac {9}{2}}x-{\frac {23}{6}})e^{3x};}
y
~
″
−
6
y
~
′
+
9
y
~
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle {\tilde {y}}''-6{\tilde {y}}'+9{\tilde {y}}=(2x-3)e^{3x},}
(
31
4
x
2
+
9
2
x
−
23
6
)
e
3
x
−
6
(
31
12
x
2
+
2
3
x
−
3
2
)
e
3
x
+
9
(
31
12
x
2
−
3
2
x
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle ({\frac {31}{4}}x^{2}+{\frac {9}{2}}x-{\frac {23}{6}})e^{3x}-6({\frac {31}{12}}x^{2}+{\frac {2}{3}}x-{\frac {3}{2}})e^{3x}+9({\frac {31}{12}}x^{2}-{\frac {3}{2}}x)e^{3x}=(2x-3)e^{3x},}
(
31
4
x
2
+
9
2
x
−
23
6
−
31
2
x
2
−
4
3
x
+
9
2
+
93
4
x
2
−
27
2
x
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle ({\frac {31}{4}}x^{2}+{\frac {9}{2}}x-{\frac {23}{6}}-{\frac {31}{2}}x^{2}-{\frac {4}{3}}x+{\frac {9}{2}}+{\frac {93}{4}}x^{2}-{\frac {27}{2}}x)e^{3x}=(2x-3)e^{3x},}
(
31
4
x
2
−
31
2
x
2
+
93
4
x
2
+
9
2
x
−
4
3
x
−
27
2
x
−
23
6
+
9
2
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle ({\frac {31}{4}}x^{2}-{\frac {31}{2}}x^{2}+{\frac {93}{4}}x^{2}+{\frac {9}{2}}x-{\frac {4}{3}}x-{\frac {27}{2}}x-{\frac {23}{6}}+{\frac {9}{2}})e^{3x}=(2x-3)e^{3x},}
(
−
31
4
x
2
+
93
4
x
2
−
4
3
x
−
18
2
x
+
54
−
46
2
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle (-{\frac {31}{4}}x^{2}+{\frac {93}{4}}x^{2}-{\frac {4}{3}}x-{\frac {18}{2}}x+{\frac {54-46}{2}})e^{3x}=(2x-3)e^{3x},}
(
62
4
x
2
−
4
3
x
−
9
x
+
8
2
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle ({\frac {62}{4}}x^{2}-{\frac {4}{3}}x-9x+{\frac {8}{2}})e^{3x}=(2x-3)e^{3x},}
(
31
2
x
2
+
−
4
−
27
3
x
+
4
)
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle ({\frac {31}{2}}x^{2}+{\frac {-4-27}{3}}x+4)e^{3x}=(2x-3)e^{3x},}
(
31
2
x
2
−
31
3
x
+
4
)
e
3
x
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle ({\frac {31}{2}}x^{2}-{\frac {31}{3}}x+4)e^{3x}=(2x-3)e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
6
y
′
+
9
y
=
0.
{\displaystyle y''-6y'+9y=0.}
k
2
−
6
k
+
9
=
0
{\displaystyle k^{2}-6k+9=0}
k
1
=
3
{\displaystyle k_{1}=3}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
¯
=
(
C
1
+
x
C
2
)
e
3
x
.
{\displaystyle {\bar {y}}=(C_{1}+xC_{2})e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
Kai
f
(
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle f(x)=(2x-3)e^{3x},}
P
n
(
x
)
=
2
x
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=2x-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
;
α
=
3
{\displaystyle Q_{n}(x)=a;\;\alpha =3}
y
~
=
x
2
a
e
3
x
=
a
x
2
e
3
x
.
{\displaystyle {\tilde {y}}=x^{2}ae^{3x}=ax^{2}e^{3x}.}
Randame išvestines:
y
~
′
=
(
a
x
2
e
3
x
)
′
=
(
3
a
x
2
+
2
a
x
)
e
3
x
,
{\displaystyle {\tilde {y}}'=(ax^{2}e^{3x})'=(3ax^{2}+2ax)e^{3x},}
y
~
″
=
(
6
a
x
+
2
a
)
e
3
x
+
3
(
3
a
x
2
+
2
a
x
)
e
3
x
=
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
.
{\displaystyle {\tilde {y}}''=(6ax+2a)e^{3x}+3(3ax^{2}+2ax)e^{3x}=(9ax^{2}+12ax+2a)e^{3x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
−
6
(
3
a
x
2
+
2
a
x
)
e
3
x
+
9
a
x
2
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle (9ax^{2}+12ax+2a)e^{3x}-6(3ax^{2}+2ax)e^{3x}+9ax^{2}e^{3x}=(2x-3)e^{3x},}
9
a
x
2
+
12
a
x
+
2
a
−
18
a
x
2
−
12
a
x
+
9
a
x
2
=
2
x
−
3
,
{\displaystyle 9ax^{2}+12ax+2a-18ax^{2}-12ax+9ax^{2}=2x-3,}
9
a
x
2
−
18
a
x
2
+
9
a
x
2
+
12
a
x
−
12
a
x
+
2
a
=
2
x
−
3
,
{\displaystyle 9ax^{2}-18ax^{2}+9ax^{2}+12ax-12ax+2a=2x-3,}
2
a
=
2
x
−
3.
{\displaystyle 2a=2x-3.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
1
|
0
=
2
,
{\displaystyle x^{1}\quad |\;0=2,}
x
0
|
2
b
=
−
3.
{\displaystyle x^{0}\quad |\;2b=-3.}
Iš sistemos randame:
a
=
0
,
b
=
−
3
/
2
=
−
1.5.
{\displaystyle a=0,\;b=-3/2=-1.5.}
y
~
=
(
0
⋅
x
−
1.5
)
e
3
x
=
−
1.5
e
3
x
.
{\displaystyle {\tilde {y}}=(0\cdot x-1.5)e^{3x}=-1.5e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}}
y
=
y
¯
+
y
~
=
C
1
e
3
x
+
x
C
2
e
3
x
−
3
2
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{3x}+xC_{2}e^{3x}-{\frac {3}{2}}e^{3x}.}
Patikriname:
y
~
′
=
(
−
3
2
e
3
x
)
′
=
−
9
2
e
3
x
,
{\displaystyle {\tilde {y}}'=(-{\frac {3}{2}}e^{3x})'=-{\frac {9}{2}}e^{3x},}
y
~
″
=
−
27
2
e
3
x
;
{\displaystyle {\tilde {y}}''=-{\frac {27}{2}}e^{3x};}
y
~
″
−
6
y
~
′
+
9
y
~
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle {\tilde {y}}''-6{\tilde {y}}'+9{\tilde {y}}=(2x-3)e^{3x},}
−
27
2
e
3
x
−
6
(
−
9
2
e
3
x
)
+
9
(
−
3
2
e
3
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}e^{3x}-6(-{\frac {9}{2}}e^{3x})+9(-{\frac {3}{2}}e^{3x})=(2x-3)e^{3x},}
−
27
2
e
3
x
+
27
e
3
x
−
27
2
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}e^{3x}+27e^{3x}-{\frac {27}{2}}e^{3x}=(2x-3)e^{3x},}
0
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle 0=(2x-3)e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
6
y
′
+
9
y
=
0.
{\displaystyle y''-6y'+9y=0.}
k
2
−
6
k
+
9
=
0
{\displaystyle k^{2}-6k+9=0}
k
1
=
3
{\displaystyle k_{1}=3}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
¯
=
(
C
1
+
x
C
2
)
e
3
x
.
{\displaystyle {\bar {y}}=(C_{1}+xC_{2})e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
Kai
f
(
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle f(x)=(2x-3)e^{3x},}
P
n
(
x
)
=
2
x
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=2x-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
;
α
=
3
{\displaystyle Q_{n}(x)=a;\;\alpha =3}
y
~
=
x
2
a
e
3
x
=
a
x
2
e
3
x
.
{\displaystyle {\tilde {y}}=x^{2}ae^{3x}=ax^{2}e^{3x}.}
Randame išvestines:
y
~
′
=
(
a
x
2
e
3
x
)
′
=
(
3
a
x
2
+
2
a
x
)
e
3
x
,
{\displaystyle {\tilde {y}}'=(ax^{2}e^{3x})'=(3ax^{2}+2ax)e^{3x},}
y
~
″
=
(
6
a
x
+
2
a
)
e
3
x
+
3
(
3
a
x
2
+
2
a
x
)
e
3
x
=
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
.
{\displaystyle {\tilde {y}}''=(6ax+2a)e^{3x}+3(3ax^{2}+2ax)e^{3x}=(9ax^{2}+12ax+2a)e^{3x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
−
6
(
3
a
x
2
+
2
a
x
)
e
3
x
+
9
a
x
2
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle (9ax^{2}+12ax+2a)e^{3x}-6(3ax^{2}+2ax)e^{3x}+9ax^{2}e^{3x}=(2x-3)e^{3x},}
9
a
x
2
+
12
a
x
+
2
a
−
18
a
x
2
−
12
a
x
+
9
a
x
2
=
2
x
−
3
,
{\displaystyle 9ax^{2}+12ax+2a-18ax^{2}-12ax+9ax^{2}=2x-3,}
9
a
x
2
−
18
a
x
2
+
9
a
x
2
+
12
a
x
−
12
a
x
+
2
a
=
2
x
−
3
,
{\displaystyle 9ax^{2}-18ax^{2}+9ax^{2}+12ax-12ax+2a=2x-3,}
2
a
=
2
x
−
3.
{\displaystyle 2a=2x-3.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
1
|
0
=
2
,
{\displaystyle x^{1}\quad |\;0=2,}
x
0
|
2
a
=
−
3.
{\displaystyle x^{0}\quad |\;2a=-3.}
Iš sistemos randame:
a
=
−
1.5.
{\displaystyle a=-1.5.}
y
~
=
−
1.5
x
e
3
x
.
{\displaystyle {\tilde {y}}=-1.5xe^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}}
y
=
y
¯
+
y
~
=
C
1
e
3
x
+
x
C
2
e
3
x
−
3
2
x
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{3x}+xC_{2}e^{3x}-{\frac {3}{2}}xe^{3x}.}
Patikriname:
y
~
′
=
(
−
3
2
x
e
3
x
)
′
=
−
9
2
x
e
3
x
−
3
2
e
3
x
,
{\displaystyle {\tilde {y}}'=(-{\frac {3}{2}}xe^{3x})'=-{\frac {9}{2}}xe^{3x}-{\frac {3}{2}}e^{3x},}
y
~
″
=
(
−
9
2
x
e
3
x
−
3
2
e
3
x
)
′
=
−
27
2
x
e
3
x
−
9
2
e
3
x
−
9
2
e
3
x
=
{\displaystyle {\tilde {y}}''=(-{\frac {9}{2}}xe^{3x}-{\frac {3}{2}}e^{3x})'=-{\frac {27}{2}}xe^{3x}-{\frac {9}{2}}e^{3x}-{\frac {9}{2}}e^{3x}=}
=
−
27
2
x
e
3
x
−
9
e
3
x
;
{\displaystyle =-{\frac {27}{2}}xe^{3x}-9e^{3x};}
y
~
″
−
6
y
~
′
+
9
y
~
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle {\tilde {y}}''-6{\tilde {y}}'+9{\tilde {y}}=(2x-3)e^{3x},}
−
27
2
x
e
3
x
−
9
e
3
x
−
6
(
−
9
2
x
e
3
x
−
3
2
e
3
x
)
+
9
(
−
3
2
x
e
3
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}xe^{3x}-9e^{3x}-6(-{\frac {9}{2}}xe^{3x}-{\frac {3}{2}}e^{3x})+9(-{\frac {3}{2}}xe^{3x})=(2x-3)e^{3x},}
−
27
2
x
e
3
x
−
9
e
3
x
+
27
x
e
3
x
+
9
e
3
x
−
27
2
x
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}xe^{3x}-9e^{3x}+27xe^{3x}+9e^{3x}-{\frac {27}{2}}xe^{3x}=(2x-3)e^{3x},}
−
27
2
x
e
3
x
+
27
x
e
3
x
−
27
2
x
e
3
x
−
9
e
3
x
+
9
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}xe^{3x}+27xe^{3x}-{\frac {27}{2}}xe^{3x}-9e^{3x}+9e^{3x}=(2x-3)e^{3x},}
0
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle 0=(2x-3)e^{3x}.}
y
″
−
6
y
′
+
9
y
=
−
3
e
3
x
.
{\displaystyle y''-6y'+9y=-3e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
6
y
′
+
9
y
=
0.
{\displaystyle y''-6y'+9y=0.}
k
2
−
6
k
+
9
=
0
{\displaystyle k^{2}-6k+9=0}
k
1
=
3
{\displaystyle k_{1}=3}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
¯
=
(
C
1
+
x
C
2
)
e
3
x
.
{\displaystyle {\bar {y}}=(C_{1}+xC_{2})e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
Kai
f
(
x
)
=
−
3
e
3
x
,
{\displaystyle f(x)=-3e^{3x},}
P
n
(
x
)
=
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
;
α
=
3
{\displaystyle Q_{n}(x)=a;\;\alpha =3}
y
~
=
x
2
a
e
3
x
=
a
x
2
e
3
x
.
{\displaystyle {\tilde {y}}=x^{2}ae^{3x}=ax^{2}e^{3x}.}
Randame išvestines:
y
~
′
=
(
a
x
2
e
3
x
)
′
=
(
3
a
x
2
+
2
a
x
)
e
3
x
,
{\displaystyle {\tilde {y}}'=(ax^{2}e^{3x})'=(3ax^{2}+2ax)e^{3x},}
y
~
″
=
(
6
a
x
+
2
a
)
e
3
x
+
3
(
3
a
x
2
+
2
a
x
)
e
3
x
=
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
.
{\displaystyle {\tilde {y}}''=(6ax+2a)e^{3x}+3(3ax^{2}+2ax)e^{3x}=(9ax^{2}+12ax+2a)e^{3x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
−
6
(
3
a
x
2
+
2
a
x
)
e
3
x
+
9
a
x
2
e
3
x
=
−
3
e
3
x
,
{\displaystyle (9ax^{2}+12ax+2a)e^{3x}-6(3ax^{2}+2ax)e^{3x}+9ax^{2}e^{3x}=-3e^{3x},}
9
a
x
2
+
12
a
x
+
2
a
−
18
a
x
2
−
12
a
x
+
9
a
x
2
=
−
3
,
{\displaystyle 9ax^{2}+12ax+2a-18ax^{2}-12ax+9ax^{2}=-3,}
9
a
x
2
−
18
a
x
2
+
9
a
x
2
+
12
a
x
−
12
a
x
+
2
a
=
−
3
,
{\displaystyle 9ax^{2}-18ax^{2}+9ax^{2}+12ax-12ax+2a=-3,}
2
a
=
−
3.
{\displaystyle 2a=-3.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname
x
0
|
2
a
=
−
3.
{\displaystyle x^{0}\quad |\;2a=-3.}
Iš sistemos randame:
a
=
−
1.5.
{\displaystyle a=-1.5.}
y
~
=
−
1.5
e
3
x
.
{\displaystyle {\tilde {y}}=-1.5e^{3x}.}
y
=
y
¯
+
y
~
=
C
1
e
3
x
+
x
C
2
e
3
x
−
3
2
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{3x}+xC_{2}e^{3x}-{\frac {3}{2}}e^{3x}.}
Patikriname:
y
~
′
=
(
−
3
2
e
3
x
)
′
=
−
9
2
e
3
x
,
{\displaystyle {\tilde {y}}'=(-{\frac {3}{2}}e^{3x})'=-{\frac {9}{2}}e^{3x},}
y
~
″
=
−
27
2
e
3
x
;
{\displaystyle {\tilde {y}}''=-{\frac {27}{2}}e^{3x};}
y
~
″
−
6
y
~
′
+
9
y
~
=
−
3
e
3
x
,
{\displaystyle {\tilde {y}}''-6{\tilde {y}}'+9{\tilde {y}}=-3e^{3x},}
−
27
2
e
3
x
−
6
(
−
9
2
e
3
x
)
+
9
(
−
3
2
e
3
x
)
=
−
3
e
3
x
,
{\displaystyle -{\frac {27}{2}}e^{3x}-6(-{\frac {9}{2}}e^{3x})+9(-{\frac {3}{2}}e^{3x})=-3e^{3x},}
−
27
2
e
3
x
+
27
e
3
x
−
27
2
e
3
x
=
−
3
e
3
x
,
{\displaystyle -{\frac {27}{2}}e^{3x}+27e^{3x}-{\frac {27}{2}}e^{3x}=-3e^{3x},}
0
=
−
3
e
3
x
.
{\displaystyle 0=-3e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}.}
Sprendimas . Pirmiausia išspręskime lygtį
y
″
−
6
y
′
+
9
y
=
0.
{\displaystyle y''-6y'+9y=0.}
k
2
−
6
k
+
9
=
0
{\displaystyle k^{2}-6k+9=0}
k
1
=
3
{\displaystyle k_{1}=3}
k
2
=
3
,
{\displaystyle k_{2}=3,}
y
¯
=
(
C
1
+
x
C
2
)
e
3
x
.
{\displaystyle {\bar {y}}=(C_{1}+xC_{2})e^{3x}.}
Toliau, atsižvelgdami į dešiniosios pusės išraišką, parinksime atskirąjį nehomogeninės lygties sprendinį.
Kai
f
(
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle f(x)=(2x-3)e^{3x},}
P
n
(
x
)
=
2
x
−
3
,
α
=
3.
{\displaystyle P_{n}(x)=2x-3,\;\alpha =3.}
P
n
(
x
)
{\displaystyle P_{n}(x)}
Q
n
(
x
)
=
a
;
α
=
3
{\displaystyle Q_{n}(x)=a;\;\alpha =3}
y
~
=
x
2
a
e
3
x
=
a
x
2
e
3
x
.
{\displaystyle {\tilde {y}}=x^{2}ae^{3x}=ax^{2}e^{3x}.}
Randame išvestines:
y
~
′
=
(
a
x
2
e
3
x
)
′
=
(
3
a
x
2
+
2
a
x
)
e
3
x
,
{\displaystyle {\tilde {y}}'=(ax^{2}e^{3x})'=(3ax^{2}+2ax)e^{3x},}
y
~
″
=
(
6
a
x
+
2
a
)
e
3
x
+
3
(
3
a
x
2
+
2
a
x
)
e
3
x
=
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
.
{\displaystyle {\tilde {y}}''=(6ax+2a)e^{3x}+3(3ax^{2}+2ax)e^{3x}=(9ax^{2}+12ax+2a)e^{3x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
(
9
a
x
2
+
12
a
x
+
2
a
)
e
3
x
−
6
(
3
a
x
2
+
2
a
x
)
e
3
x
+
9
a
x
2
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle (9ax^{2}+12ax+2a)e^{3x}-6(3ax^{2}+2ax)e^{3x}+9ax^{2}e^{3x}=(2x-3)e^{3x},}
9
a
x
2
+
12
a
x
+
2
a
−
18
a
x
2
−
12
a
x
+
9
a
x
2
=
2
x
−
3
,
{\displaystyle 9ax^{2}+12ax+2a-18ax^{2}-12ax+9ax^{2}=2x-3,}
9
a
x
2
−
18
a
x
2
+
9
a
x
2
+
12
a
x
−
12
a
x
+
2
a
=
2
x
−
3
,
{\displaystyle 9ax^{2}-18ax^{2}+9ax^{2}+12ax-12ax+2a=2x-3,}
2
a
=
2
x
−
3.
{\displaystyle 2a=2x-3.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
1
|
0
=
2
,
{\displaystyle x^{1}\quad |\;0=2,}
x
0
|
2
a
=
−
3.
{\displaystyle x^{0}\quad |\;2a=-3.}
Iš sistemos randame:
a
=
−
1.5.
{\displaystyle a=-1.5.}
y
~
=
−
1.5
x
2
e
3
x
.
{\displaystyle {\tilde {y}}=-1.5x^{2}e^{3x}.}
y
″
−
6
y
′
+
9
y
=
(
2
x
−
3
)
e
3
x
{\displaystyle y''-6y'+9y=(2x-3)e^{3x}}
y
=
y
¯
+
y
~
=
C
1
e
3
x
+
x
C
2
e
3
x
−
3
2
x
2
e
3
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=C_{1}e^{3x}+xC_{2}e^{3x}-{\frac {3}{2}}x^{2}e^{3x}.}
Patikriname:
y
~
′
=
(
−
3
2
x
2
e
3
x
)
′
=
−
9
2
x
2
e
3
x
−
6
2
x
e
3
x
,
{\displaystyle {\tilde {y}}'=(-{\frac {3}{2}}x^{2}e^{3x})'=-{\frac {9}{2}}x^{2}e^{3x}-{\frac {6}{2}}xe^{3x},}
y
~
″
=
(
−
9
2
x
2
e
3
x
−
6
2
x
e
3
x
)
′
=
−
18
2
x
e
3
x
−
6
2
e
3
x
+
3
(
−
9
2
x
2
−
6
2
x
)
e
3
x
=
{\displaystyle {\tilde {y}}''=(-{\frac {9}{2}}x^{2}e^{3x}-{\frac {6}{2}}xe^{3x})'=-{\frac {18}{2}}xe^{3x}-{\frac {6}{2}}e^{3x}+3(-{\frac {9}{2}}x^{2}-{\frac {6}{2}}x)e^{3x}=}
=
−
9
x
e
3
x
−
3
e
3
x
−
27
2
x
2
e
3
x
−
9
x
e
3
x
=
−
27
2
x
2
e
3
x
−
18
x
e
3
x
−
3
e
3
x
;
{\displaystyle =-9xe^{3x}-3e^{3x}-{\frac {27}{2}}x^{2}e^{3x}-9xe^{3x}=-{\frac {27}{2}}x^{2}e^{3x}-18xe^{3x}-3e^{3x};}
y
~
″
−
6
y
~
′
+
9
y
~
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle {\tilde {y}}''-6{\tilde {y}}'+9{\tilde {y}}=(2x-3)e^{3x},}
−
27
2
x
2
e
3
x
−
18
x
e
3
x
−
3
e
3
x
−
6
(
−
9
2
x
2
e
3
x
−
6
2
x
e
3
x
)
+
9
(
−
3
2
x
2
e
3
x
)
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}x^{2}e^{3x}-18xe^{3x}-3e^{3x}-6(-{\frac {9}{2}}x^{2}e^{3x}-{\frac {6}{2}}xe^{3x})+9(-{\frac {3}{2}}x^{2}e^{3x})=(2x-3)e^{3x},}
−
27
2
x
2
e
3
x
−
18
x
e
3
x
−
3
e
3
x
+
27
x
2
e
3
x
+
18
x
e
3
x
−
27
2
x
2
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -{\frac {27}{2}}x^{2}e^{3x}-18xe^{3x}-3e^{3x}+27x^{2}e^{3x}+18xe^{3x}-{\frac {27}{2}}x^{2}e^{3x}=(2x-3)e^{3x},}
−
18
x
e
3
x
−
3
e
3
x
+
18
x
e
3
x
=
(
2
x
−
3
)
e
3
x
,
{\displaystyle -18xe^{3x}-3e^{3x}+18xe^{3x}=(2x-3)e^{3x},}
−
3
e
3
x
=
(
2
x
−
3
)
e
3
x
.
{\displaystyle -3e^{3x}=(2x-3)e^{3x}.}
y
″
−
4
y
′
+
13
y
=
f
(
x
)
,
{\displaystyle y''-4y'+13y=f(x),}
kai: a)
f
(
x
)
=
x
e
−
2
x
;
{\displaystyle f(x)=xe^{-2x};}
f
(
x
)
=
e
−
2
x
sin
(
3
x
)
.
{\displaystyle f(x)=e^{-2x}\sin(3x).}
Sprendimas . Kadangi charakteringoji lygtis
k
2
+
4
k
+
13
=
0
{\displaystyle k^{2}+4k+13=0}
k
1
,
2
=
−
2
±
3
i
,
{\displaystyle k_{1,2}=-2\pm 3i,}
y
″
−
4
y
′
+
13
y
=
0
{\displaystyle y''-4y'+13y=0}
y
¯
=
e
−
2
x
(
C
1
cos
(
3
x
)
+
C
2
sin
(
3
x
)
)
.
{\displaystyle {\bar {y}}=e^{-2x}(C_{1}\cos(3x)+C_{2}\sin(3x)).}
a) Kai
f
(
x
)
=
x
e
−
2
x
,
{\displaystyle f(x)=xe^{-2x},}
P
n
(
x
)
e
α
x
.
{\displaystyle P_{n}(x)e^{\alpha x}.}
Kadangi šį kartą
P
n
(
x
)
=
x
{\displaystyle P_{n}(x)=x}
α
=
−
2
{\displaystyle \alpha =-2}
α
=
−
2
{\displaystyle \alpha =-2}
−
2
±
3
i
{\displaystyle -2\pm 3i}
y
~
{\displaystyle {\tilde {y}}}
y
~
=
(
a
x
+
b
)
e
−
2
x
.
{\displaystyle {\tilde {y}}=(ax+b)e^{-2x}.}
Randame išvestines:
y
~
′
=
(
(
a
x
+
b
)
e
−
2
x
)
′
=
a
e
−
2
x
−
2
(
a
x
+
b
)
e
−
2
x
=
(
−
2
a
x
+
a
−
b
)
e
−
2
x
,
{\displaystyle {\tilde {y}}'=((ax+b)e^{-2x})'=ae^{-2x}-2(ax+b)e^{-2x}=(-2ax+a-b)e^{-2x},}
y
~
″
=
−
2
a
e
−
2
x
−
2
(
−
2
a
x
+
a
−
b
)
e
−
2
x
=
−
2
a
e
−
2
x
+
(
4
a
x
−
2
a
+
2
b
)
e
−
2
x
=
2
(
2
a
x
−
a
+
b
)
e
−
2
x
.
{\displaystyle {\tilde {y}}''=-2ae^{-2x}-2(-2ax+a-b)e^{-2x}=-2ae^{-2x}+(4ax-2a+2b)e^{-2x}=2(2ax-a+b)e^{-2x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
y
~
″
−
4
y
~
′
+
13
y
~
=
x
e
−
2
x
,
{\displaystyle {\tilde {y}}''-4{\tilde {y}}'+13{\tilde {y}}=xe^{-2x},}
2
(
2
a
x
−
a
+
b
)
e
−
2
x
−
4
(
−
2
a
x
+
a
−
b
)
e
−
2
x
+
13
(
a
x
+
b
)
e
−
2
x
=
x
e
−
2
x
,
{\displaystyle 2(2ax-a+b)e^{-2x}-4(-2ax+a-b)e^{-2x}+13(ax+b)e^{-2x}=xe^{-2x},}
4
a
x
−
2
a
+
2
b
+
8
a
x
−
4
a
+
4
b
+
13
a
x
+
13
b
=
x
,
{\displaystyle 4ax-2a+2b+8ax-4a+4b+13ax+13b=x,}
25
a
x
−
6
a
+
19
b
=
x
.
{\displaystyle 25ax-6a+19b=x.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
1
|
25
a
=
1
,
{\displaystyle x^{1}\quad |\;25a=1,}
x
0
|
−
6
a
+
19
b
=
0.
{\displaystyle x^{0}\quad |\;-6a+19b=0.}
Iš sistemos randame:
a
=
1
25
=
0.04
,
b
=
6
a
19
=
6
25
⋅
19
=
6
475
=
0.012631579.
{\displaystyle a={\frac {1}{25}}=0.04,\;b={\frac {6a}{19}}={\frac {6}{25\cdot 19}}={\frac {6}{475}}=0.012631579.}
y
~
=
(
x
25
+
6
475
)
e
−
2
x
.
{\displaystyle {\tilde {y}}=\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}.}
y
=
y
¯
+
y
~
=
e
−
2
x
(
C
1
cos
(
3
x
)
+
C
2
sin
(
3
x
)
)
+
(
x
25
+
6
475
)
e
−
2
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=e^{-2x}(C_{1}\cos(3x)+C_{2}\sin(3x))+\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}.}
Patikriname:
y
~
′
=
(
(
x
25
+
6
475
)
e
−
2
x
)
′
=
1
25
e
−
2
x
−
2
(
x
25
+
6
475
)
e
−
2
x
,
{\displaystyle {\tilde {y}}'=\left(\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}\right)'={\frac {1}{25}}e^{-2x}-2\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x},}
=
(
−
2
25
x
+
1
25
−
12
475
)
e
−
2
x
=
(
−
2
25
x
+
19
−
12
475
)
e
−
2
x
=
(
−
2
25
x
+
7
475
)
e
−
2
x
,
{\displaystyle =\left(-{\frac {2}{25}}x+{\frac {1}{25}}-{\frac {12}{475}}\right)e^{-2x}=\left(-{\frac {2}{25}}x+{\frac {19-12}{475}}\right)e^{-2x}=\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x},}
y
~
″
=
−
2
25
e
−
2
x
−
2
(
−
2
25
x
+
7
475
)
e
−
2
x
=
(
4
25
x
−
14
475
−
2
25
)
e
−
2
x
=
{\displaystyle {\tilde {y}}''=-{\frac {2}{25}}e^{-2x}-2\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x}=\left({\frac {4}{25}}x-{\frac {14}{475}}-{\frac {2}{25}}\right)e^{-2x}=}
=
(
4
25
x
−
14
+
36
475
)
e
−
2
x
=
(
4
25
x
−
50
475
)
e
−
2
x
;
{\displaystyle =\left({\frac {4}{25}}x-{\frac {14+36}{475}}\right)e^{-2x}=\left({\frac {4}{25}}x-{\frac {50}{475}}\right)e^{-2x};}
y
~
″
−
4
y
~
′
+
13
y
~
=
x
e
−
2
x
,
{\displaystyle {\tilde {y}}''-4{\tilde {y}}'+13{\tilde {y}}=xe^{-2x},}
(
4
25
x
−
50
475
)
e
−
2
x
−
4
(
−
2
25
x
+
7
475
)
e
−
2
x
+
13
(
x
25
+
6
475
)
e
−
2
x
=
x
e
−
2
x
,
{\displaystyle \left({\frac {4}{25}}x-{\frac {50}{475}}\right)e^{-2x}-4\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x}+13\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}=xe^{-2x},}
4
25
x
−
50
475
+
8
25
x
−
28
475
+
13
25
x
+
78
475
=
x
,
{\displaystyle {\frac {4}{25}}x-{\frac {50}{475}}+{\frac {8}{25}}x-{\frac {28}{475}}+{\frac {13}{25}}x+{\frac {78}{475}}=x,}
4
+
8
+
13
25
x
+
78
−
50
−
28
475
=
x
,
{\displaystyle {\frac {4+8+13}{25}}x+{\frac {78-50-28}{475}}=x,}
25
25
x
+
0
475
=
x
,
{\displaystyle {\frac {25}{25}}x+{\frac {0}{475}}=x,}
x
=
x
.
{\displaystyle x=x.}
y
″
−
4
y
′
+
13
y
=
x
e
−
2
x
.
{\displaystyle y''-4y'+13y=xe^{-2x}.}
Sprendimas . Kad rasti homogeninės lygties
y
″
−
4
y
′
+
13
y
=
0
{\displaystyle y''-4y'+13y=0}
y
=
e
k
x
{\displaystyle y=e^{kx}}
(
e
k
x
)
″
−
4
(
e
k
x
)
′
+
13
e
k
x
=
0
,
{\displaystyle (e^{kx})''-4(e^{kx})'+13e^{kx}=0,}
k
2
e
k
x
−
4
k
e
k
x
+
13
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}-4ke^{kx}+13e^{kx}=0,}
k
2
−
4
k
+
13
=
0.
{\displaystyle k^{2}-4k+13=0.}
Kadangi charakteringoji lygtis
k
2
−
4
k
+
13
=
0
{\displaystyle k^{2}-4k+13=0}
šaknis
k
1
,
2
=
2
±
3
i
,
{\displaystyle k_{1,2}=2\pm 3i,}
y
″
−
4
y
′
+
13
y
=
0
{\displaystyle y''-4y'+13y=0}
y
¯
=
e
2
x
(
C
1
cos
(
3
x
)
+
C
2
sin
(
3
x
)
)
.
{\displaystyle {\bar {y}}=e^{2x}(C_{1}\cos(3x)+C_{2}\sin(3x)).}
Kai
f
(
x
)
=
x
e
−
2
x
,
{\displaystyle f(x)=xe^{-2x},}
P
n
(
x
)
e
α
x
.
{\displaystyle P_{n}(x)e^{\alpha x}.}
Kadangi
P
n
(
x
)
=
x
{\displaystyle P_{n}(x)=x}
α
=
−
2
{\displaystyle \alpha =-2}
y
~
{\displaystyle {\tilde {y}}}
y
~
=
(
a
x
+
b
)
e
−
2
x
.
{\displaystyle {\tilde {y}}=(ax+b)e^{-2x}.}
Randame išvestines:
y
~
′
=
(
(
a
x
+
b
)
e
−
2
x
)
′
=
a
e
−
2
x
−
2
(
a
x
+
b
)
e
−
2
x
=
(
−
2
a
x
+
a
−
b
)
e
−
2
x
,
{\displaystyle {\tilde {y}}'=((ax+b)e^{-2x})'=ae^{-2x}-2(ax+b)e^{-2x}=(-2ax+a-b)e^{-2x},}
y
~
″
=
−
2
a
e
−
2
x
−
2
(
−
2
a
x
+
a
−
b
)
e
−
2
x
=
−
2
a
e
−
2
x
+
(
4
a
x
−
2
a
+
2
b
)
e
−
2
x
=
2
(
2
a
x
−
a
+
b
)
e
−
2
x
.
{\displaystyle {\tilde {y}}''=-2ae^{-2x}-2(-2ax+a-b)e^{-2x}=-2ae^{-2x}+(4ax-2a+2b)e^{-2x}=2(2ax-a+b)e^{-2x}.}
Įrašę
y
~
,
y
~
′
,
y
~
″
{\displaystyle {\tilde {y}},\;{\tilde {y}}',\;{\tilde {y}}''}
y
~
″
−
4
y
~
′
+
13
y
~
=
x
e
−
2
x
,
{\displaystyle {\tilde {y}}''-4{\tilde {y}}'+13{\tilde {y}}=xe^{-2x},}
2
(
2
a
x
−
a
+
b
)
e
−
2
x
−
4
(
−
2
a
x
+
a
−
b
)
e
−
2
x
+
13
(
a
x
+
b
)
e
−
2
x
=
x
e
−
2
x
,
{\displaystyle 2(2ax-a+b)e^{-2x}-4(-2ax+a-b)e^{-2x}+13(ax+b)e^{-2x}=xe^{-2x},}
4
a
x
−
2
a
+
2
b
+
8
a
x
−
4
a
+
4
b
+
13
a
x
+
13
b
=
x
,
{\displaystyle 4ax-2a+2b+8ax-4a+4b+13ax+13b=x,}
25
a
x
−
6
a
+
19
b
=
x
.
{\displaystyle 25ax-6a+19b=x.}
Sulyginę koeficientus prie vienodų x laipsnių, gauname sistemą
x
1
|
25
a
=
1
,
{\displaystyle x^{1}\quad |\;25a=1,}
x
0
|
−
6
a
+
19
b
=
0.
{\displaystyle x^{0}\quad |\;-6a+19b=0.}
Iš sistemos randame:
a
=
1
25
=
0.04
,
b
=
6
a
19
=
6
25
⋅
19
=
6
475
=
0.012631579.
{\displaystyle a={\frac {1}{25}}=0.04,\;b={\frac {6a}{19}}={\frac {6}{25\cdot 19}}={\frac {6}{475}}=0.012631579.}
y
~
=
(
x
25
+
6
475
)
e
−
2
x
.
{\displaystyle {\tilde {y}}=\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}.}
y
=
y
¯
+
y
~
=
e
2
x
(
C
1
cos
(
3
x
)
+
C
2
sin
(
3
x
)
)
+
(
x
25
+
6
475
)
e
−
2
x
.
{\displaystyle y={\bar {y}}+{\tilde {y}}=e^{2x}(C_{1}\cos(3x)+C_{2}\sin(3x))+\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}.}
Patikriname:
y
~
′
=
(
(
x
25
+
6
475
)
e
−
2
x
)
′
=
1
25
e
−
2
x
−
2
(
x
25
+
6
475
)
e
−
2
x
,
{\displaystyle {\tilde {y}}'=\left(\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}\right)'={\frac {1}{25}}e^{-2x}-2\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x},}
=
(
−
2
25
x
+
1
25
−
12
475
)
e
−
2
x
=
(
−
2
25
x
+
19
−
12
475
)
e
−
2
x
=
(
−
2
25
x
+
7
475
)
e
−
2
x
,
{\displaystyle =\left(-{\frac {2}{25}}x+{\frac {1}{25}}-{\frac {12}{475}}\right)e^{-2x}=\left(-{\frac {2}{25}}x+{\frac {19-12}{475}}\right)e^{-2x}=\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x},}
y
~
″
=
−
2
25
e
−
2
x
−
2
(
−
2
25
x
+
7
475
)
e
−
2
x
=
(
4
25
x
−
14
475
−
2
25
)
e
−
2
x
=
{\displaystyle {\tilde {y}}''=-{\frac {2}{25}}e^{-2x}-2\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x}=\left({\frac {4}{25}}x-{\frac {14}{475}}-{\frac {2}{25}}\right)e^{-2x}=}
=
(
4
25
x
−
14
+
36
475
)
e
−
2
x
=
(
4
25
x
−
50
475
)
e
−
2
x
;
{\displaystyle =\left({\frac {4}{25}}x-{\frac {14+36}{475}}\right)e^{-2x}=\left({\frac {4}{25}}x-{\frac {50}{475}}\right)e^{-2x};}
y
~
″
−
4
y
~
′
+
13
y
~
=
x
e
−
2
x
,
{\displaystyle {\tilde {y}}''-4{\tilde {y}}'+13{\tilde {y}}=xe^{-2x},}
(
4
25
x
−
50
475
)
e
−
2
x
−
4
(
−
2
25
x
+
7
475
)
e
−
2
x
+
13
(
x
25
+
6
475
)
e
−
2
x
=
x
e
−
2
x
,
{\displaystyle \left({\frac {4}{25}}x-{\frac {50}{475}}\right)e^{-2x}-4\left(-{\frac {2}{25}}x+{\frac {7}{475}}\right)e^{-2x}+13\left({\frac {x}{25}}+{\frac {6}{475}}\right)e^{-2x}=xe^{-2x},}
4
25
x
−
50
475
+
8
25
x
−
28
475
+
13
25
x
+
78
475
=
x
,
{\displaystyle {\frac {4}{25}}x-{\frac {50}{475}}+{\frac {8}{25}}x-{\frac {28}{475}}+{\frac {13}{25}}x+{\frac {78}{475}}=x,}
4
+
8
+
13
25
x
+
78
−
50
−
28
475
=
x
,
{\displaystyle {\frac {4+8+13}{25}}x+{\frac {78-50-28}{475}}=x,}
25
25
x
+
0
475
=
x
,
{\displaystyle {\frac {25}{25}}x+{\frac {0}{475}}=x,}
x
=
x
.
{\displaystyle x=x.}
Grįžti į "Matematika/Antrosios eilės tiesinės nehomogeninės diferencialinės lygtys" puslapį.
