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Matematika/Bernulio diferencialinė lygtis
Kalba
Stebėti
Keisti
<
Matematika
(Nukreipta iš
Bernulio diferencialinė lygtis
)
Bernulio diferencialinė lygtis
y
′
+
P
(
x
)
y
=
Q
(
x
)
y
m
,
{\displaystyle y'+P(x)y=Q(x)y^{m},}
y
−
m
y
′
+
P
(
x
)
y
1
−
m
=
Q
(
x
)
,
{\displaystyle y^{-m}y'+P(x)y^{1-m}=Q(x),}
y
1
−
m
=
z
,
{\displaystyle y^{1-m}=z,\;}
z
′
=
(
1
−
m
)
y
−
m
y
′
,
{\displaystyle z'=(1-m)y^{-m}y',\;}
z
′
1
−
m
=
y
−
m
y
′
,
{\displaystyle {\frac {z'}{1-m}}=y^{-m}y',}
z
′
1
−
m
+
P
(
x
)
z
=
Q
(
x
)
,
{\displaystyle {\frac {z'}{1-m}}+P(x)z=Q(x),}
z
′
+
(
1
−
m
)
P
(
x
)
z
=
(
1
−
m
)
Q
(
x
)
.
{\displaystyle z'+(1-m)P(x)z=(1-m)Q(x).}
Bernulio lygtį galima spręsti panašiai kaip ir pirmosios eilės tiesinę, naudojant keitinį
y
=
u
v
.
{\displaystyle y=uv.}
x
y
′
+
y
+
x
2
y
2
e
x
=
0
,
{\displaystyle xy'+y+x^{2}y^{2}e^{x}=0,}
y
′
+
y
x
=
−
x
y
2
e
x
;
{\displaystyle y'+{y \over x}=-xy^{2}e^{x};}
y
=
u
v
,
{\displaystyle y=uv,}
y
′
=
u
′
v
+
u
v
′
{\displaystyle y'=u'v+uv'}
,
u
′
v
+
u
v
′
+
u
v
x
=
−
x
u
2
v
2
e
x
,
{\displaystyle u'v+uv'+{uv \over x}=-xu^{2}v^{2}e^{x},}
v
(
u
′
+
u
x
)
+
u
v
′
=
−
x
u
2
v
2
e
x
;
{\displaystyle v(u'+{u \over x})+uv'=-xu^{2}v^{2}e^{x};}
u
′
+
u
x
=
0
,
{\displaystyle u'+{u \over x}=0,}
d
u
d
x
=
−
u
x
,
{\displaystyle {du \over dx}=-{u \over x},}
∫
d
u
u
=
−
∫
d
x
x
,
{\displaystyle \int {du \over u}=-\int {dx \over x},}
ln
|
u
|
=
−
ln
|
x
|
,
{\displaystyle \ln |u|=-\ln |x|,}
u
=
1
x
;
{\displaystyle u={1 \over x};}
1
x
v
′
=
−
x
⋅
1
x
2
v
2
e
x
,
{\displaystyle {1 \over x}v'=-x\cdot {1 \over x^{2}}v^{2}e^{x},}
v
′
=
−
v
2
e
x
,
{\displaystyle v'=-v^{2}e^{x},}
∫
d
v
v
2
=
−
∫
e
x
d
x
,
{\displaystyle \int {dv \over v^{2}}=-\int e^{x}dx,}
−
1
v
=
−
(
e
x
+
C
)
,
{\displaystyle -{1 \over v}=-(e^{x}+C),}
v
=
1
e
x
+
C
;
{\displaystyle v={1 \over e^{x}+C};}
y
=
u
v
=
1
x
(
e
x
+
C
)
.
{\displaystyle y=uv={1 \over x(e^{x}+C)}.}
d
y
d
x
−
3
x
y
=
−
x
3
y
2
,
{\displaystyle {dy \over dx}-{3 \over x}y=-x^{3}y^{2},}
1
y
2
d
y
d
x
−
3
x
1
y
=
−
x
3
,
{\displaystyle {1 \over y^{2}}{dy \over dx}-{3 \over x}{1 \over y}=-x^{3},}
1
y
=
z
,
−
1
y
2
d
y
d
x
=
d
z
d
x
,
{\displaystyle {1 \over y}=z,\;-{1 \over y^{2}}{dy \over dx}={dz \over dx},}
d
z
d
x
+
3
x
z
=
x
3
;
{\displaystyle {dz \over dx}+{3 \over x}z=x^{3};}
šią tiesinę lygtį integruosime konstantos variacijos metodu,
d
z
d
x
+
3
x
z
=
0
,
{\displaystyle {dz \over dx}+{3 \over x}z=0,}
∫
d
z
z
=
−
3
∫
d
x
x
,
{\displaystyle \int {dz \over z}=-3\int {dx \over x},}
ln
|
z
|
=
−
3
ln
|
x
|
+
ln
|
C
|
=
ln
|
C
x
−
3
|
,
{\displaystyle \ln |z|=-3\ln |x|+\ln |C|=\ln |Cx^{-3}|,}
z
=
C
x
3
;
{\displaystyle z={C \over x^{3}};}
C
=
C
(
x
)
,
{\displaystyle C=C(x),}
z
=
C
(
x
)
x
−
3
,
{\displaystyle z=C(x)x^{-3},}
d
z
d
x
=
d
C
(
x
)
d
x
1
x
3
−
3
C
(
x
)
x
4
,
{\displaystyle {dz \over dx}={dC(x) \over dx}{1 \over x^{3}}-{3C(x) \over x^{4}},}
d
z
d
x
+
3
x
z
=
d
C
(
x
)
d
x
1
x
3
−
3
C
(
x
)
x
4
+
3
x
C
(
x
)
x
3
=
x
3
,
{\displaystyle {dz \over dx}+{3 \over x}z={dC(x) \over dx}{1 \over x^{3}}-{3C(x) \over x^{4}}+{3 \over x}{C(x) \over x^{3}}=x^{3},}
d
C
(
x
)
d
x
1
x
3
=
x
3
,
{\displaystyle {dC(x) \over dx}{1 \over x^{3}}=x^{3},}
d
C
(
x
)
d
x
=
x
6
,
{\displaystyle {dC(x) \over dx}=x^{6},}
∫
d
C
(
x
)
=
∫
x
6
d
x
,
{\displaystyle \int dC(x)=\int x^{6}dx,}
C
(
x
)
=
x
7
7
+
C
1
;
{\displaystyle C(x)={x^{7} \over 7}+C_{1};}
z
=
C
(
x
)
x
3
=
x
7
7
+
C
1
x
3
=
x
4
7
+
C
1
x
3
,
{\displaystyle z={C(x) \over x^{3}}={{x^{7} \over 7}+C_{1} \over x^{3}}={x^{4} \over 7}+{C_{1} \over x^{3}},}
z
=
1
y
,
1
y
=
x
4
7
+
C
1
x
3
,
{\displaystyle z={1 \over y},\;{1 \over y}={x^{4} \over 7}+{C_{1} \over x^{3}},}
y
=
1
x
4
7
+
C
1
x
3
=
1
x
7
+
7
C
1
7
x
3
=
7
x
3
x
7
+
7
C
1
.
{\displaystyle y={1 \over {x^{4} \over 7}+{C_{1} \over x^{3}}}={1 \over {x^{7}+7C_{1} \over 7x^{3}}}={7x^{3} \over x^{7}+7C_{1}}.}
,
