Antrosios eilės tiesine homogenine diferencialine lygtimi su pastoviaisiais koeficientais vadinama lygtis
y
″
+
p
y
′
+
q
y
=
0.
{\displaystyle y''+py'+qy=0.}
Tokia lygtis išsprendžiama parinkus
y
=
e
k
x
.
{\displaystyle y=e^{kx}.}
Toliau gauname,
(
e
k
x
)
″
+
p
(
e
k
x
)
′
+
q
e
k
x
=
0
,
{\displaystyle (e^{kx})''+p(e^{kx})'+qe^{kx}=0,}
k
2
e
k
x
+
k
p
e
k
x
+
q
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}+kpe^{kx}+qe^{kx}=0,}
(
k
2
+
k
p
+
q
)
e
k
x
=
0.
{\displaystyle (k^{2}+kp+q)e^{kx}=0.}
Kadangi nėra tokio k , kad
e
k
x
{\displaystyle e^{kx}}
k
2
+
k
p
+
q
=
0.
{\displaystyle k^{2}+kp+q=0.}
Išsprendžiant šią lygtį ir bus gautas diferencialinės lygties sprendinys (arba du sprendiniai). Yra trys atvejai, kai
k
1
≠
k
2
,
{\displaystyle k_{1}\neq k_{2},}
k
1
=
k
2
{\displaystyle k_{1}=k_{2}}
Vronskio determinantas pavadintas Juzefo Vronskio (J. Wronski, 1776-1853) vardu. Oficialiai laikoma, kad Vronskio determinantas padeda spręsti Antrosios eilės tiesines homogenines diferencialines lygtis su pastoviaisiais koeficientais , tačiau teisybė yra, kad jis neturi su jomis nieko bendro (senovėje šito nesuprato arba ir dabar ne visi supranta, todėl neišima jo iš vadovelių). Dėl šios priežasties Vronskio determinantas čia nebus nagrinėjamas. Bent jau Vronskio determinantas tikrai nereikalingas kai
k
1
=
k
2
.
{\displaystyle k_{1}=k_{2}.}
Vronskio determinanto radimas:
W
(
y
1
,
y
2
)
=
[
y
1
y
2
y
1
′
y
2
′
]
{\displaystyle W(y_{1},y_{2})={\begin{bmatrix}y_{1}&&y_{2}\\y'_{1}&&y'_{2}\end{bmatrix}}}
Kai
y
1
{\displaystyle y_{1}}
y
2
{\displaystyle y_{2}}
k
1
{\displaystyle k_{1}}
k
2
{\displaystyle k_{2}}
keisti
Jeigu lygties
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
sprendiniai
k
1
{\displaystyle k_{1}}
k
2
{\displaystyle k_{2}}
y
1
=
e
k
1
x
,
y
2
=
e
k
2
x
.
{\displaystyle y_{1}=e^{k_{1}x},\quad y_{2}=e^{k_{2}x}.}
Bendrasis lygties
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
sprendinys yra
y
=
C
1
e
k
1
x
+
C
2
e
k
2
x
.
{\displaystyle y=C_{1}e^{k_{1}x}+C_{2}e^{k_{2}x}.}
Įrodymas . Turime lygtį
y
″
+
p
y
′
+
q
y
=
0.
{\displaystyle y''+py'+qy=0.}
Tokia lygtis išsprendžiama parinkus
y
=
e
k
x
.
{\displaystyle y=e^{kx}.}
Toliau gauname,
(
e
k
x
)
″
+
p
(
e
k
x
)
′
+
q
e
k
x
=
0
,
{\displaystyle (e^{kx})''+p(e^{kx})'+qe^{kx}=0,}
k
2
e
k
x
+
k
p
e
k
x
+
q
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}+kpe^{kx}+qe^{kx}=0,}
(
k
2
+
k
p
+
q
)
e
k
x
=
0.
{\displaystyle (k^{2}+kp+q)e^{kx}=0.}
Kadangi nėra tokio k , kad
e
k
x
{\displaystyle e^{kx}}
k
2
+
k
p
+
q
=
0.
{\displaystyle k^{2}+kp+q=0.}
Gavome, kad
k
1
≠
k
2
.
{\displaystyle k_{1}\neq k_{2}.}
y
1
=
e
k
1
x
{\displaystyle y_{1}=e^{k_{1}x}}
y
2
=
e
k
2
x
.
{\displaystyle y_{2}=e^{k_{2}x}.}
Bet mes tikimes, kad gali būti ir sudetingesni sprendiniai, todėl parenkame tokią funkciją z , kuri gali buti arba funkcija nuo x (kaip pavyzdžiui
z
=
C
x
+
C
1
{\displaystyle z=Cx+C_{1}}
z
=
C
{\displaystyle z=C}
z padauginame su diferencialinės lygties sprendiniais ir gauname:
y
10
=
z
e
k
1
x
{\displaystyle y_{10}=ze^{k_{1}x}}
y
20
=
z
e
k
2
x
.
{\displaystyle y_{20}=ze^{k_{2}x}.}
Randame
y
10
{\displaystyle y_{10}}
y
20
{\displaystyle y_{20}}
y
10
′
=
(
z
e
k
1
x
)
′
=
z
′
e
k
1
x
+
z
k
1
e
k
1
x
,
{\displaystyle y_{10}'=(ze^{k_{1}x})'=z'e^{k_{1}x}+zk_{1}e^{k_{1}x},}
y
10
″
=
(
z
′
e
k
1
x
+
z
k
1
e
k
1
x
)
′
=
(
z
″
e
k
1
x
+
z
′
k
1
e
k
1
x
)
+
(
z
′
k
1
e
k
1
x
+
z
k
1
2
e
k
1
x
)
,
{\displaystyle y_{10}''=(z'e^{k_{1}x}+zk_{1}e^{k_{1}x})'=(z''e^{k_{1}x}+z'k_{1}e^{k_{1}x})+(z'k_{1}e^{k_{1}x}+zk_{1}^{2}e^{k_{1}x}),}
y
10
″
=
z
″
e
k
1
x
+
2
k
1
z
′
e
k
1
x
+
k
1
2
z
e
k
1
x
;
{\displaystyle y_{10}''=z''e^{k_{1}x}+2k_{1}z'e^{k_{1}x}+k_{1}^{2}ze^{k_{1}x};}
y
20
′
=
(
z
e
k
2
x
)
′
=
z
′
e
k
2
x
+
k
2
z
e
k
1
x
,
{\displaystyle y_{20}'=(ze^{k_{2}x})'=z'e^{k_{2}x}+k_{2}ze^{k_{1}x},}
y
20
″
=
z
″
e
k
2
x
+
2
k
2
z
′
e
k
2
x
+
k
2
2
z
e
k
2
x
.
{\displaystyle y_{20}''=z''e^{k_{2}x}+2k_{2}z'e^{k_{2}x}+k_{2}^{2}ze^{k_{2}x}.}
Įstatome
y
10
,
y
10
′
,
y
10
″
{\displaystyle y_{10},\;y_{10}',\;y_{10}''\;}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
z
″
e
k
1
x
+
2
k
1
z
′
e
k
1
x
+
k
1
2
z
e
k
1
x
+
p
(
z
′
e
k
1
x
+
k
1
z
e
k
1
x
)
+
q
z
e
k
1
x
=
0
,
{\displaystyle z''e^{k_{1}x}+2k_{1}z'e^{k_{1}x}+k_{1}^{2}ze^{k_{1}x}+p(z'e^{k_{1}x}+k_{1}ze^{k_{1}x})+qze^{k_{1}x}=0,}
z
″
e
k
1
x
+
(
2
k
1
+
p
)
z
′
e
k
1
x
+
(
k
1
2
+
p
k
1
+
q
)
z
e
k
1
x
=
0
,
{\displaystyle z''e^{k_{1}x}+(2k_{1}+p)z'e^{k_{1}x}+(k_{1}^{2}+pk_{1}+q)ze^{k_{1}x}=0,}
z
″
+
(
2
k
1
+
p
)
z
′
+
(
k
1
2
+
p
k
1
+
q
)
z
=
0.
{\displaystyle z''+(2k_{1}+p)z'+(k_{1}^{2}+pk_{1}+q)z=0.}
Mums reikia, kad
z
″
{\displaystyle z''}
z
′
=
0
{\displaystyle z'=0}
k
1
2
+
p
k
1
+
q
=
0
{\displaystyle k_{1}^{2}+pk_{1}+q=0}
k
1
{\displaystyle k_{1}}
z turi būti konstanta, nes tik tada
z
″
=
0
{\displaystyle z''=0}
z
′
=
0
{\displaystyle z'=0}
z
=
C
1
{\displaystyle z=C_{1}}
C
1
″
+
(
2
k
1
+
p
)
C
1
′
+
(
k
1
2
+
p
k
1
+
q
)
C
1
=
0
,
{\displaystyle C_{1}''+(2k_{1}+p)C_{1}'+(k_{1}^{2}+pk_{1}+q)C_{1}=0,}
(
k
1
2
+
p
k
1
+
q
)
C
1
=
0.
{\displaystyle (k_{1}^{2}+pk_{1}+q)C_{1}=0.}
Šis paskutinis reiškinys tikrai lygus nuliui su bet kokiu skaičiumi
C
1
{\displaystyle C_{1}}
k
1
{\displaystyle k_{1}}
Analogiškai randame, kad
(
k
2
2
+
p
k
2
+
q
)
C
2
=
0.
{\displaystyle (k_{2}^{2}+pk_{2}+q)C_{2}=0.}
Vadinasi,
y
10
=
C
1
e
k
1
x
,
{\displaystyle y_{10}=C_{1}e^{k_{1}x},}
y
20
=
C
2
e
k
2
x
.
{\displaystyle y_{20}=C_{2}e^{k_{2}x}.}
Suprantame, kad tiek su
C
1
e
k
1
x
{\displaystyle C_{1}e^{k_{1}x}}
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
C
2
e
k
2
x
{\displaystyle C_{2}e^{k_{2}x}}
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
(
f
(
x
)
+
g
(
x
)
)
′
=
f
′
(
x
)
+
g
′
(
x
)
,
{\displaystyle (f(x)+g(x))'=f'(x)+g'(x),\;}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
y
10
+
y
20
=
C
1
e
k
1
x
+
C
2
e
k
2
x
,
{\displaystyle y=y_{10}+y_{20}=C_{1}e^{k_{1}x}+C_{2}e^{k_{2}x},}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
C
1
e
k
1
x
+
C
2
e
k
2
x
.
{\displaystyle y=C_{1}e^{k_{1}x}+C_{2}e^{k_{2}x}.}
Išspręskime lygtį
y
″
+
y
′
−
6
y
=
0.
{\displaystyle y''+y'-6y=0.}
Sprendimas . Įstatome vietoje y reikšmę
e
k
x
{\displaystyle e^{kx}}
(
e
k
x
)
″
+
(
e
k
x
)
′
−
6
e
k
x
=
0
,
{\displaystyle (e^{kx})''+(e^{kx})'-6e^{kx}=0,}
k
2
e
k
x
+
k
e
k
x
−
6
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}+ke^{kx}-6e^{kx}=0,}
padalinus abi puses iš
e
k
x
{\displaystyle e^{kx}}
k
2
+
k
−
6
=
0
;
{\displaystyle k^{2}+k-6=0;}
D
=
b
2
−
4
a
c
=
1
2
−
4
⋅
1
⋅
(
−
6
)
=
1
+
24
=
25
;
{\displaystyle D=b^{2}-4ac=1^{2}-4\cdot 1\cdot (-6)=1+24=25;}
k
1
=
−
b
−
D
2
a
=
−
1
−
25
2
⋅
1
=
−
6
2
=
−
3
;
{\displaystyle k_{1}={\frac {-b-{\sqrt {D}}}{2a}}={\frac {-1-{\sqrt {25}}}{2\cdot 1}}={\frac {-6}{2}}=-3;}
k
2
=
−
b
+
D
2
a
=
−
1
+
25
2
⋅
1
=
4
2
=
2.
{\displaystyle k_{2}={\frac {-b+{\sqrt {D}}}{2a}}={\frac {-1+{\sqrt {25}}}{2\cdot 1}}={\frac {4}{2}}=2.}
Charakteringoji lygtis
y
″
+
y
′
−
6
y
=
0
{\displaystyle y''+y'-6y=0}
k
1
=
−
3
{\displaystyle k_{1}=-3}
k
2
=
2.
{\displaystyle k_{2}=2.}
y
=
C
1
e
−
3
x
+
C
2
e
2
x
.
{\displaystyle y=C_{1}e^{-3x}+C_{2}e^{2x}.}
Patikriname, kad
y
=
C
1
e
−
3
x
+
C
2
e
2
x
{\displaystyle y=C_{1}e^{-3x}+C_{2}e^{2x}}
y
″
+
y
′
−
6
y
=
0
{\displaystyle y''+y'-6y=0}
y
′
=
−
3
C
1
e
−
3
x
+
2
C
2
e
2
x
,
{\displaystyle y'=-3C_{1}e^{-3x}+2C_{2}e^{2x},}
y
″
=
9
C
1
e
−
3
x
+
4
C
2
e
2
x
;
{\displaystyle y''=9C_{1}e^{-3x}+4C_{2}e^{2x};}
(
9
C
1
e
−
3
x
+
4
C
2
e
2
x
)
+
(
−
3
C
1
e
−
3
x
+
2
C
2
e
2
x
)
−
6
(
C
1
e
−
3
x
+
C
2
e
2
x
)
=
0
,
{\displaystyle (9C_{1}e^{-3x}+4C_{2}e^{2x})+(-3C_{1}e^{-3x}+2C_{2}e^{2x})-6(C_{1}e^{-3x}+C_{2}e^{2x})=0,}
9
C
1
e
−
3
x
+
4
C
2
e
2
x
−
3
C
1
e
−
3
x
+
2
C
2
e
2
x
−
6
C
1
e
−
3
x
−
6
C
2
e
2
x
=
0
,
{\displaystyle 9C_{1}e^{-3x}+4C_{2}e^{2x}-3C_{1}e^{-3x}+2C_{2}e^{2x}-6C_{1}e^{-3x}-6C_{2}e^{2x}=0,}
9
C
1
e
−
3
x
+
6
C
2
e
2
x
−
9
C
1
e
−
3
x
−
6
C
2
e
2
x
=
0.
{\displaystyle 9C_{1}e^{-3x}+6C_{2}e^{2x}-9C_{1}e^{-3x}-6C_{2}e^{2x}=0.}
k
1
{\displaystyle k_{1}}
k
2
{\displaystyle k_{2}}
keisti
Kadangi šiuo atveju
e
k
1
x
=
e
k
2
x
,
{\displaystyle e^{k_{1}x}=e^{k_{2}x},}
y
1
=
e
k
x
.
{\displaystyle y_{1}=e^{kx}.}
y
2
=
z
(
x
)
e
k
x
.
{\displaystyle y_{2}=z(x)e^{kx}.}
Tuomet
y
2
′
=
(
z
(
x
)
e
k
x
)
′
=
z
′
(
x
)
e
k
x
+
k
z
(
x
)
e
k
x
,
{\displaystyle y_{2}'=(z(x)e^{kx})'=z'(x)e^{kx}+kz(x)e^{kx},\;}
y
2
″
=
(
z
(
x
)
e
k
x
)
″
=
z
″
(
x
)
e
k
x
+
k
z
′
(
x
)
e
k
x
+
k
z
′
(
x
)
e
k
x
+
k
2
z
(
x
)
e
k
x
,
{\displaystyle y_{2}''=(z(x)e^{kx})''=z''(x)e^{kx}+kz'(x)e^{kx}+kz'(x)e^{kx}+k^{2}z(x)e^{kx},}
y
2
″
=
(
z
(
x
)
e
k
x
)
″
=
z
″
(
x
)
e
k
x
+
2
k
z
′
(
x
)
e
k
x
+
k
2
z
(
x
)
e
k
x
.
{\displaystyle y_{2}''=(z(x)e^{kx})''=z''(x)e^{kx}+2kz'(x)e^{kx}+k^{2}z(x)e^{kx}.\;}
Įrašę
y
2
,
{\displaystyle y_{2},}
y
2
′
{\displaystyle y_{2}'}
y
2
″
{\displaystyle y_{2}''}
y
″
+
p
y
′
+
q
y
=
0
,
{\displaystyle y''+py'+qy=0,}
z
″
(
x
)
e
k
x
+
2
k
z
′
(
x
)
e
k
x
+
k
2
z
(
x
)
e
k
x
+
p
(
z
′
(
x
)
e
k
x
+
k
z
(
x
)
e
k
x
)
+
q
z
(
x
)
e
k
x
=
0
,
{\displaystyle z''(x)e^{kx}+2kz'(x)e^{kx}+k^{2}z(x)e^{kx}+p(z'(x)e^{kx}+kz(x)e^{kx})+qz(x)e^{kx}=0,}
z
″
(
x
)
e
k
x
+
(
2
k
+
p
)
z
′
(
x
)
e
k
x
+
(
k
2
+
p
k
+
q
)
z
(
x
)
e
k
x
=
0
,
{\displaystyle z''(x)e^{kx}+(2k+p)z'(x)e^{kx}+(k^{2}+pk+q)z(x)e^{kx}=0,}
z
″
(
x
)
+
(
2
k
+
p
)
z
′
(
x
)
+
(
k
2
+
p
k
+
q
)
z
(
x
)
=
0.
{\displaystyle z''(x)+(2k+p)z'(x)+(k^{2}+pk+q)z(x)=0.}
Pagal Vieto teoremą
p
=
−
(
k
1
+
k
2
)
=
−
(
k
+
k
)
=
−
2
k
,
{\displaystyle p=-(k_{1}+k_{2})=-(k+k)=-2k,}
2
k
=
−
p
{\displaystyle 2k=-p}
2
k
+
p
=
0.
{\displaystyle 2k+p=0.}
z
″
(
x
)
+
0
⋅
z
′
(
x
)
+
(
k
2
+
p
k
+
q
)
z
(
x
)
=
0
,
{\displaystyle z''(x)+0\cdot z'(x)+(k^{2}+pk+q)z(x)=0,}
z
″
(
x
)
+
(
k
2
+
p
k
+
q
)
z
(
x
)
=
0.
{\displaystyle z''(x)+(k^{2}+pk+q)z(x)=0.}
Kad gauti nulį ir kaip nors išspręsti šitą lygtį mums reikia, kad reikšmė
z
″
(
x
)
{\displaystyle z''(x)}
(
k
2
+
p
k
+
q
)
z
(
x
)
=
0
,
{\displaystyle (k^{2}+pk+q)z(x)=0,}
kurią mes jau galime išspresti reikalaudami, kad
k
2
+
p
k
+
q
=
0
{\displaystyle k^{2}+pk+q=0}
Matome, kad
z
″
(
x
)
=
0
,
{\displaystyle z''(x)=0,}
z
(
x
)
=
A
x
+
A
1
,
{\displaystyle z(x)=Ax+A_{1},}
z
′
(
x
)
=
(
A
x
+
A
1
)
′
=
A
,
{\displaystyle z'(x)=(Ax+A_{1})'=A,}
z
″
(
x
)
=
A
′
=
0.
{\displaystyle z''(x)=A'=0.}
Todėl turime bendrąjį sprendinį
y
=
y
2
=
z
(
x
)
e
k
x
=
(
A
x
+
A
1
)
e
k
x
,
{\displaystyle y=y_{2}=z(x)e^{kx}=(Ax+A_{1})e^{kx},}
kurį galime užrašyti taip:
y
=
C
1
e
k
x
+
C
2
x
e
k
x
=
e
k
x
(
C
1
+
x
C
2
)
.
{\displaystyle y=C_{1}e^{kx}+C_{2}xe^{kx}=e^{kx}(C_{1}+xC_{2}).}
Išspręskime lygtį
y
″
+
8
y
′
+
16
y
=
0.
{\displaystyle y''+8y'+16y=0.}
Sprendimas . Įstatę
y
=
e
k
x
{\displaystyle y=e^{kx}}
k
2
e
k
x
+
8
k
e
k
x
+
16
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}+8ke^{kx}+16e^{kx}=0,}
k
2
+
8
k
+
16
=
0.
{\displaystyle k^{2}+8k+16=0.}
Charakteringoji lygtis
k
2
+
8
k
+
16
=
0
{\displaystyle k^{2}+8k+16=0}
k
1
=
k
2
=
−
4
{\displaystyle k_{1}=k_{2}=-4}
k
1
⋅
k
2
=
−
4
⋅
(
−
4
)
=
16
=
q
{\displaystyle k_{1}\cdot k_{2}=-4\cdot (-4)=16=q}
p
=
−
(
k
1
+
k
2
)
=
−
(
−
4
−
4
)
=
8
{\displaystyle p=-(k_{1}+k_{2})=-(-4-4)=8}
D
=
8
2
−
4
⋅
16
=
0
{\displaystyle D=8^{2}-4\cdot 16=0}
y
=
e
−
4
x
(
C
1
+
x
C
2
)
.
{\displaystyle y=e^{-4x}(C_{1}+xC_{2}).}
Patikriname:
y
′
=
(
C
1
e
−
4
x
+
C
2
x
e
−
4
x
)
′
=
−
4
C
1
e
−
4
x
+
C
2
e
−
4
x
−
4
C
2
x
e
−
4
x
;
{\displaystyle y'=(C_{1}e^{-4x}+C_{2}xe^{-4x})'=-4C_{1}e^{-4x}+C_{2}e^{-4x}-4C_{2}xe^{-4x};}
y
″
=
(
−
4
C
1
e
−
4
x
+
C
2
e
−
4
x
−
4
C
2
x
e
−
4
x
)
′
=
16
C
1
e
−
4
x
−
4
C
2
e
−
4
x
−
4
C
2
e
−
4
x
+
16
C
2
x
e
−
4
x
,
{\displaystyle y''=(-4C_{1}e^{-4x}+C_{2}e^{-4x}-4C_{2}xe^{-4x})'=16C_{1}e^{-4x}-4C_{2}e^{-4x}-4C_{2}e^{-4x}+16C_{2}xe^{-4x},}
y
″
=
16
C
1
e
−
4
x
−
8
C
2
e
−
4
x
+
16
C
2
x
e
−
4
x
.
{\displaystyle y''=16C_{1}e^{-4x}-8C_{2}e^{-4x}+16C_{2}xe^{-4x}.}
Įstatome
y
{\displaystyle y}
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
y
″
+
8
y
′
+
16
y
=
0
{\displaystyle y''+8y'+16y=0}
16
C
1
e
−
4
x
−
8
C
2
e
−
4
x
+
16
C
2
x
e
−
4
x
+
8
(
−
4
C
1
e
−
4
x
+
C
2
e
−
4
x
−
4
C
2
x
e
−
4
x
)
+
16
(
C
1
e
−
4
x
+
C
2
x
e
−
4
x
)
=
0
,
{\displaystyle 16C_{1}e^{-4x}-8C_{2}e^{-4x}+16C_{2}xe^{-4x}+8(-4C_{1}e^{-4x}+C_{2}e^{-4x}-4C_{2}xe^{-4x})+16(C_{1}e^{-4x}+C_{2}xe^{-4x})=0,}
16
C
1
e
−
4
x
−
8
C
2
e
−
4
x
+
16
C
2
x
e
−
4
x
−
32
C
1
e
−
4
x
+
8
C
2
e
−
4
x
−
32
C
2
x
e
−
4
x
+
16
C
1
e
−
4
x
+
16
C
2
x
e
−
4
x
=
0
,
{\displaystyle 16C_{1}e^{-4x}-8C_{2}e^{-4x}+16C_{2}xe^{-4x}-32C_{1}e^{-4x}+8C_{2}e^{-4x}-32C_{2}xe^{-4x}+16C_{1}e^{-4x}+16C_{2}xe^{-4x}=0,}
16
C
1
e
−
4
x
−
32
C
1
e
−
4
x
+
16
C
1
e
−
4
x
−
8
C
2
e
−
4
x
+
8
C
2
e
−
4
x
+
16
C
2
x
e
−
4
x
−
32
C
2
x
e
−
4
x
+
16
C
2
x
e
−
4
x
=
0
,
{\displaystyle 16C_{1}e^{-4x}-32C_{1}e^{-4x}+16C_{1}e^{-4x}-8C_{2}e^{-4x}+8C_{2}e^{-4x}+16C_{2}xe^{-4x}-32C_{2}xe^{-4x}+16C_{2}xe^{-4x}=0,}
(
16
C
1
e
−
4
x
−
32
C
1
e
−
4
x
+
16
C
1
e
−
4
x
)
+
(
−
8
C
2
e
−
4
x
+
8
C
2
e
−
4
x
)
+
(
16
C
2
x
e
−
4
x
−
32
C
2
x
e
−
4
x
+
16
C
2
x
e
−
4
x
)
=
0.
{\displaystyle (16C_{1}e^{-4x}-32C_{1}e^{-4x}+16C_{1}e^{-4x})+(-8C_{2}e^{-4x}+8C_{2}e^{-4x})+(16C_{2}xe^{-4x}-32C_{2}xe^{-4x}+16C_{2}xe^{-4x})=0.}
k
1
{\displaystyle k_{1}}
k
2
{\displaystyle k_{2}}
keisti
k
1
=
α
+
β
i
,
{\displaystyle k_{1}=\alpha +\beta i,}
k
2
=
α
−
β
i
;
{\displaystyle k_{2}=\alpha -\beta i;}
čia
α
=
−
p
2
,
β
=
q
−
p
2
4
,
{\displaystyle \alpha =-{\frac {p}{2}},\;\beta ={\sqrt {q-{\frac {p^{2}}{4}}}},}
q
−
p
2
4
>
0.
{\displaystyle q-{\frac {p^{2}}{4}}>0.}
Įrodyta, kad tokiu atveju bendrasis lygties
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
e
α
x
(
C
1
cos
(
β
x
)
+
C
2
sin
(
β
x
)
)
.
{\displaystyle y=e^{\alpha x}(C_{1}\cos(\beta x)+C_{2}\sin(\beta x)).}
Įrodymas . Į lygtį
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
e
k
x
{\displaystyle y=e^{kx}}
(
e
k
x
)
″
+
p
(
e
k
x
)
′
+
q
e
k
x
=
0
,
{\displaystyle (e^{kx})''+p(e^{kx})'+qe^{kx}=0,}
k
2
e
k
x
+
k
p
e
k
x
+
q
e
k
x
=
0
,
{\displaystyle k^{2}e^{kx}+kpe^{kx}+qe^{kx}=0,}
k
2
+
k
p
+
q
=
0
;
{\displaystyle k^{2}+kp+q=0;}
D
=
p
2
−
4
q
,
(
D
<
0
)
{\displaystyle D=p^{2}-4q,\quad (D<0)}
k
1
=
−
p
+
D
2
=
−
p
+
p
2
−
4
q
2
=
−
p
+
i
2
(
4
q
−
p
2
)
2
=
−
p
+
i
4
q
−
p
2
2
=
−
p
2
+
i
4
q
−
p
2
4
,
{\displaystyle k_{1}={\frac {-p+{\sqrt {D}}}{2}}={\frac {-p+{\sqrt {p^{2}-4q}}}{2}}={\frac {-p+{\sqrt {i^{2}(4q-p^{2})}}}{2}}={\frac {-p+i{\sqrt {4q-p^{2}}}}{2}}=-{\frac {p}{2}}+i{\sqrt {\frac {4q-p^{2}}{4}}},}
k
2
=
−
p
−
D
2
=
−
p
−
p
2
−
4
q
2
=
−
p
−
i
2
(
4
q
−
p
2
)
2
=
−
p
−
i
4
q
−
p
2
2
=
−
p
2
−
i
4
q
−
p
2
4
.
{\displaystyle k_{2}={\frac {-p-{\sqrt {D}}}{2}}={\frac {-p-{\sqrt {p^{2}-4q}}}{2}}={\frac {-p-{\sqrt {i^{2}(4q-p^{2})}}}{2}}={\frac {-p-i{\sqrt {4q-p^{2}}}}{2}}=-{\frac {p}{2}}-i{\sqrt {\frac {4q-p^{2}}{4}}}.}
Taigi, gauname du sprendinius
k
1
=
α
+
i
β
{\displaystyle k_{1}=\alpha +i\beta }
k
2
=
α
−
i
β
.
{\displaystyle k_{2}=\alpha -i\beta .}
Nesunku suprasti, kad jeigu vieną sprendinį (pavyzdžiui,
k
1
=
α
+
i
β
{\displaystyle k_{1}=\alpha +i\beta }
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
k
2
=
α
−
i
β
{\displaystyle k_{2}=\alpha -i\beta }
y
=
e
k
1
x
=
e
(
α
+
i
β
)
x
{\displaystyle y=e^{k_{1}x}=e^{(\alpha +i\beta )x}}
y
″
+
p
y
′
+
q
y
=
0
,
{\displaystyle y''+py'+qy=0,}
(
e
(
α
+
i
β
)
x
)
″
+
p
(
e
(
α
+
i
β
)
x
)
′
+
q
e
(
α
+
i
β
)
x
=
0
,
{\displaystyle (e^{(\alpha +i\beta )x})''+p(e^{(\alpha +i\beta )x})'+qe^{(\alpha +i\beta )x}=0,}
(
α
+
i
β
)
2
e
(
α
+
i
β
)
x
+
(
α
+
i
β
)
p
e
(
α
+
i
β
)
x
+
q
e
(
α
+
i
β
)
x
=
0
,
{\displaystyle (\alpha +i\beta )^{2}e^{(\alpha +i\beta )x}+(\alpha +i\beta )pe^{(\alpha +i\beta )x}+qe^{(\alpha +i\beta )x}=0,}
(
−
p
2
+
i
4
q
−
p
2
4
)
2
+
(
−
p
2
+
i
4
q
−
p
2
4
)
p
+
q
=
0
,
{\displaystyle (-{\frac {p}{2}}+i{\sqrt {\frac {4q-p^{2}}{4}}})^{2}+(-{\frac {p}{2}}+i{\sqrt {\frac {4q-p^{2}}{4}}})p+q=0,}
p
2
4
−
2
⋅
p
2
⋅
i
4
q
−
p
2
4
−
4
q
−
p
2
4
+
(
−
p
2
+
i
4
q
−
p
2
4
)
p
+
q
=
0
,
{\displaystyle {\frac {p^{2}}{4}}-2\cdot {\frac {p}{2}}\cdot i{\sqrt {\frac {4q-p^{2}}{4}}}-{\frac {4q-p^{2}}{4}}+(-{\frac {p}{2}}+i{\sqrt {\frac {4q-p^{2}}{4}}})p+q=0,}
p
2
4
−
i
p
4
q
−
p
2
4
−
4
q
−
p
2
4
−
p
2
2
+
i
p
4
q
−
p
2
4
+
q
=
0
,
{\displaystyle {\frac {p^{2}}{4}}-ip{\sqrt {\frac {4q-p^{2}}{4}}}-{\frac {4q-p^{2}}{4}}-{\frac {p^{2}}{2}}+ip{\sqrt {\frac {4q-p^{2}}{4}}}+q=0,}
p
2
4
−
4
q
−
p
2
4
−
p
2
2
+
q
=
0
,
{\displaystyle {\frac {p^{2}}{4}}-{\frac {4q-p^{2}}{4}}-{\frac {p^{2}}{2}}+q=0,}
−
4
q
−
p
2
4
−
p
2
4
+
q
=
0
,
{\displaystyle -{\frac {4q-p^{2}}{4}}-{\frac {p^{2}}{4}}+q=0,}
−
q
+
p
2
4
−
p
2
4
+
q
=
0.
{\displaystyle -q+{\frac {p^{2}}{4}}-{\frac {p^{2}}{4}}+q=0.}
Abiejose lygybės pusėse gausime nulius, taip pat įstačius
y
=
e
k
2
x
=
e
(
α
−
i
β
)
x
{\displaystyle y=e^{k_{2}x}=e^{(\alpha -i\beta )x}}
y
″
+
p
y
′
+
q
y
=
0.
{\displaystyle y''+py'+qy=0.}
Be abejonės gausime abiejose lygties
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
C
1
e
k
1
x
=
C
1
e
(
α
+
i
β
)
x
,
{\displaystyle y=C_{1}e^{k_{1}x}=C_{1}e^{(\alpha +i\beta )x},}
(
C
f
(
x
)
)
″
+
(
C
f
(
x
)
)
′
+
(
C
f
(
x
)
)
=
C
f
″
(
x
)
+
C
f
′
(
x
)
+
C
f
(
x
)
.
{\displaystyle (Cf(x))''+(Cf(x))'+(Cf(x))=Cf''(x)+Cf'(x)+Cf(x).\;}
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
y
=
C
2
e
k
2
x
=
C
2
e
(
α
−
i
β
)
x
.
{\displaystyle y=C_{2}e^{k_{2}x}=C_{2}e^{(\alpha -i\beta )x}.}
Nesunku suprasti, kad į reiškinį
y
″
+
p
y
′
+
q
y
{\displaystyle y''+py'+qy}
y
=
e
k
1
x
+
e
k
2
x
=
e
(
α
+
i
β
)
x
+
e
(
α
−
i
β
)
x
{\displaystyle y=e^{k_{1}x}+e^{k_{2}x}=e^{(\alpha +i\beta )x}+e^{(\alpha -i\beta )x}}
(
f
(
x
)
+
g
(
x
)
)
′
=
f
′
(
x
)
+
g
′
(
x
)
.
{\displaystyle (f(x)+g(x))'=f'(x)+g'(x).\;}
y
=
C
1
e
(
α
+
i
β
)
x
+
C
2
e
(
α
−
i
β
)
x
{\displaystyle y=C_{1}e^{(\alpha +i\beta )x}+C_{2}e^{(\alpha -i\beta )x}}
y
″
+
p
y
′
+
q
y
.
{\displaystyle y''+py'+qy.}
Kad atsikratyti i užrašykime sprendinį taip:
y
10
=
C
1
e
α
x
+
i
β
x
+
C
2
e
α
x
−
i
β
x
=
e
α
x
(
C
1
e
i
β
x
+
C
2
e
−
i
β
x
)
.
{\displaystyle y_{10}=C_{1}e^{\alpha x+i\beta x}+C_{2}e^{\alpha x-i\beta x}=e^{\alpha x}(C_{1}e^{i\beta x}+C_{2}e^{-i\beta x}).\;}
Toliau iš trigonometrijos ir kompleksinių skaičių žinome, kad
cos
x
=
e
i
x
+
e
−
i
x
2
{\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}}
sin
x
=
e
i
x
−
e
−
i
x
2
i
.
{\displaystyle \sin x={\frac {e^{ix}-e^{-ix}}{2i}}.}
C
1
=
1
2
{\displaystyle C_{1}={\frac {1}{2}}}
C
2
=
1
2
{\displaystyle C_{2}={\frac {1}{2}}}
y
11
=
1
2
e
α
x
+
i
β
x
+
1
2
e
α
x
−
i
β
x
=
e
α
x
e
i
β
x
+
e
−
i
β
x
2
=
e
α
x
cos
(
β
x
)
.
{\displaystyle y_{11}={\frac {1}{2}}e^{\alpha x+i\beta x}+{\frac {1}{2}}e^{\alpha x-i\beta x}=e^{\alpha x}{\frac {e^{i\beta x}+e^{-i\beta x}}{2}}=e^{\alpha x}\cos(\beta x).}
Toliau parinkime
C
1
=
1
2
i
{\displaystyle C_{1}={\frac {1}{2i}}}
C
2
=
−
1
2
i
{\displaystyle C_{2}=-{\frac {1}{2i}}}
y
12
=
1
2
i
e
α
x
+
i
β
x
−
1
2
i
e
α
x
−
i
β
x
=
e
α
x
e
i
β
x
−
e
−
i
β
x
2
i
=
e
α
x
sin
(
β
x
)
.
{\displaystyle y_{12}={\frac {1}{2i}}e^{\alpha x+i\beta x}-{\frac {1}{2i}}e^{\alpha x-i\beta x}=e^{\alpha x}{\frac {e^{i\beta x}-e^{-i\beta x}}{2i}}=e^{\alpha x}\sin(\beta x).}
Vėl žinome, kad jeigu sprendinys
y
11
=
e
α
x
cos
(
β
x
)
{\displaystyle y_{11}=e^{\alpha x}\cos(\beta x)}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
12
=
e
α
x
sin
(
β
x
)
{\displaystyle y_{12}=e^{\alpha x}\sin(\beta x)}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
20
=
y
11
+
y
12
=
e
α
x
cos
(
β
x
)
+
e
α
x
sin
(
β
x
)
{\displaystyle y_{20}=y_{11}+y_{12}=e^{\alpha x}\cos(\beta x)+e^{\alpha x}\sin(\beta x)}
y
″
+
p
y
′
+
q
y
=
0
,
{\displaystyle y''+py'+qy=0,}
(
y
11
+
y
12
)
′
=
y
11
′
+
y
12
′
.
{\displaystyle (y_{11}+y_{12})'=y_{11}'+y_{12}'.}
y
21
=
C
1
y
11
+
C
2
y
12
{\displaystyle y_{21}=C_{1}y_{11}+C_{2}y_{12}}
y
″
+
p
y
′
+
q
y
=
0
{\displaystyle y''+py'+qy=0}
y
=
C
1
e
α
x
cos
(
β
x
)
+
C
2
e
α
x
sin
(
β
x
)
=
e
α
x
(
C
1
cos
(
β
x
)
+
C
2
sin
(
β
x
)
)
.
{\displaystyle y=C_{1}e^{\alpha x}\cos(\beta x)+C_{2}e^{\alpha x}\sin(\beta x)=e^{\alpha x}(C_{1}\cos(\beta x)+C_{2}\sin(\beta x)).\;}
y
″
−
8
y
′
+
25
y
=
0.
{\displaystyle y''-8y'+25y=0.}
Sprendimas . Charakteringoji lygtis
k
2
−
8
k
+
25
=
0
{\displaystyle k^{2}-8k+25=0}
k
1
,
2
=
4
±
3
i
.
{\displaystyle k_{1,2}=4\pm 3i.}
α
=
4
,
β
=
3.
{\displaystyle \alpha =4,\;\beta =3.}
y
=
e
α
x
(
C
1
cos
(
β
x
)
+
C
2
sin
(
β
x
)
)
,
{\displaystyle y=e^{\alpha x}(C_{1}\cos(\beta x)+C_{2}\sin(\beta x)),}
y
=
e
4
x
(
C
1
cos
(
3
x
)
+
C
2
sin
(
3
x
)
)
.
{\displaystyle y=e^{4x}(C_{1}\cos(3x)+C_{2}\sin(3x)).}
Čia
α
=
−
p
2
=
−
−
8
2
=
4
,
{\displaystyle \alpha =-{\frac {p}{2}}=-{\frac {-8}{2}}=4,}
β
=
q
−
p
2
4
=
25
−
(
−
8
)
2
4
=
25
−
64
4
=
25
−
16
=
9
=
3.
{\displaystyle \beta ={\sqrt {q-{\frac {p^{2}}{4}}}}={\sqrt {25-{\frac {(-8)^{2}}{4}}}}={\sqrt {25-{\frac {64}{4}}}}={\sqrt {25-16}}={\sqrt {9}}=3.}
