Изводи
и релации
1. Да се покаже дека функцијата
y=Asin(wt+wo)+Bcos(wt+wo)y=Asin(wt+wo)+Bcos(wt+wo) size 12{y=A"sin" \( ital "wt"+ ital "wo" \) +B"cos" \( ital "wt"+ ital "wo" \) `} {},
(A,B,w,wo−const)(A,B,w,wo−const) size 12{` \( A,B,w, ital "wo" - ital "const" \) `} {}, ја задоволува релацијата:
d2ydt2+w2y=0d2ydt2+w2y=0 size 12{ { {d rSup { size 8{2} } y} over { ital "dt" rSup { size 8{2} } } } +w rSup { size 8{2} } y=0} {}.
Решение.
y
=
A
sin
(
wt
+
wo
)
+
B
cos
(
wt
+
wo
)
y
=
A
sin
(
wt
+
wo
)
+
B
cos
(
wt
+
wo
)
size 12{y=A"sin" \( ital "wt"+ ital "wo" \) +B"cos" \( ital "wt"+ ital "wo" \) } {}
(1)
dy
dt
=
A
cos
(
wt
+
wo
)
w
−
B
sin
(
wt
+
wo
)
w
dy
dt
=
A
cos
(
wt
+
wo
)
w
−
B
sin
(
wt
+
wo
)
w
size 12{ { { ital "dy"} over { ital "dt"} } =A"cos" \( ital "wt"+ ital "wo" \) w - B"sin" \( ital "wt"+ ital "wo" \) w} {}
(2)
d
2
y
dt
2
=
−
A
sin
(
wt
+
wo
)
w
2
−
B
cos
(
wt
+
wo
)
w
2
d
2
y
dt
2
=
−
A
sin
(
wt
+
wo
)
w
2
−
B
cos
(
wt
+
wo
)
w
2
size 12{ { {d rSup { size 8{2} } y} over { ital "dt" rSup { size 8{2} } } } = - A"sin" \( ital "wt"+ ital "wo" \) w rSup { size 8{2} } - B"cos" \( ital "wt"+ ital "wo" \) w rSup { size 8{2} } } {}
(3)
d
2
y
dt
2
+
w
2
y
=
−
Aw
2
sin
(
wt
+
wo
)
−
Bw
2
cos
(
wt
+
wo
)
+
w
2
(
A
sin
(
wt
+
wo
)
+
B
sin
(
wt
+
wo
)
)
=
0
d
2
y
dt
2
+
w
2
y
=
−
Aw
2
sin
(
wt
+
wo
)
−
Bw
2
cos
(
wt
+
wo
)
+
w
2
(
A
sin
(
wt
+
wo
)
+
B
sin
(
wt
+
wo
)
)
=
0
alignl { stack {
size 12{ { {d rSup { size 8{2} } y} over { ital "dt" rSup { size 8{2} } } } +w rSup { size 8{2} } y={}} {} #
- ital "Aw" rSup { size 8{2} } "sin" \( ital "wt"+ ital "wo" \) - ital "Bw" rSup { size 8{2} } "cos" \( ital "wt"+ ital "wo" \) +w rSup { size 8{2} } \( A"sin" \( ital "wt"+ ital "wo" \) +B"sin" \( ital "wt"+ ital "wo" \) \) =0 {}
} } {}
(4)2. Да се покаже дека функцијата
a1enx+a2e−nx+a3cosnx+a4sinnxa1enx+a2e−nx+a3cosnx+a4sinnx size 12{a rSub { size 8{1} } e rSup { size 8{ ital "nx"} } +a rSub { size 8{2} } e rSup { size 8{ - ital "nx"} } +a rSub { size 8{3} } "cos" ital "nx"+a rSub { size 8{4} } "sin" ital "nx"} {} ,
(a1,a2,a3,a4,n−const)(a1,a2,a3,a4,n−const) size 12{ \( a"" lSub { size 8{1} } ,a rSub { size 8{2} } ,a rSub { size 8{3} } ,a rSub { size 8{4} } ,n - ital "const" \) } {}, ја задоволува релацијата:
d4ydx4=n4yd4ydx4=n4y size 12{ { {d rSup { size 8{4} } y} over { ital "dx" rSup { size 8{4} } } } =n rSup { size 8{4} } y} {}.
Решение.
dy
dx
=
na
1
e
nx
−
a
2
ne
−
nx
−
a
3
n
sin
nx
+
a
4
n
cos
nx
dy
dx
=
na
1
e
nx
−
a
2
ne
−
nx
−
a
3
n
sin
nx
+
a
4
n
cos
nx
size 12{ { { ital "dy"} over { ital "dx"} } = ital "na" rSub { size 8{1} } e rSup { size 8{ ital "nx"} } - a rSub { size 8{2} } ital "ne" rSup { size 8{ - ital "nx"} } - a rSub { size 8{3} } n"sin" ital "nx"+a rSub { size 8{4} } n"cos" ital "nx"} {}
(5)
d
2
y
dx
2
=
n
2
a
1
e
nx
+
a
2
n
2
e
−
nx
−
n
2
a
3
cos
nx
−
a
4
n
2
sin
nx
d
2
y
dx
2
=
n
2
a
1
e
nx
+
a
2
n
2
e
−
nx
−
n
2
a
3
cos
nx
−
a
4
n
2
sin
nx
size 12{ { {d rSup { size 8{2} } y} over { ital "dx" rSup { size 8{2} } } } =n rSup { size 8{2} } a rSub { size 8{1} } e rSup { size 8{ ital "nx"} } +a rSub { size 8{2} } n rSup { size 8{2} } e rSup { size 8{ - ital "nx"} } - n rSup { size 8{2} } a rSub { size 8{3} } "cos" ital "nx" - a rSub { size 8{4} } n rSup { size 8{2} } "sin" ital "nx"} {}
d
3
y
dx
3
=
n
3
a
1
e
nx
−
a
2
n
4
e
−
nx
+
a
3
n
3
sin
nx
−
a
4
n
3
cos
nx
d
3
y
dx
3
=
n
3
a
1
e
nx
−
a
2
n
4
e
−
nx
+
a
3
n
3
sin
nx
−
a
4
n
3
cos
nx
size 12{ { {d rSup { size 8{3} } y} over { ital "dx" rSup { size 8{3} } } } =n rSup { size 8{3} } a rSub { size 8{1} } e rSup { size 8{ ital "nx"} } - a rSub { size 8{2} } n rSup { size 8{4} } e rSup { size 8{ - ital "nx"} } +a rSub { size 8{3} } n rSup { size 8{3} } "sin" ital "nx" - a rSub { size 8{4} } n rSup { size 8{3} } "cos" ital "nx"} {}
(6)
d
4
y
dx
4
=
n
4
a
1
e
nx
+
a
2
n
4
e
−
nx
+
a
3
n
4
cos
nx
+
a
4
n
4
sin
nx
d
4
y
dx
4
=
n
4
a
1
e
nx
+
a
2
n
4
e
−
nx
+
a
3
n
4
cos
nx
+
a
4
n
4
sin
nx
size 12{ { {d rSup { size 8{4} } y} over { ital "dx" rSup { size 8{4} } } } =n rSup { size 8{4} } a rSub { size 8{1} } e rSup { size 8{ ital "nx"} } +a rSub { size 8{2} } n rSup { size 8{4} } e rSup { size 8{ - ital "nx"} } +a rSub { size 8{3} } n rSup { size 8{4} } "cos" ital "nx"+a rSub { size 8{4} } n rSup { size 8{4} } "sin" ital "nx"} {}
(7)
d
4
y
dx
4
=
n
4
a
1
e
nx
+
a
2
e
−
nx
+
a
3
cos
nx
+
a
4
sin
nx
=
n
4
y
d
4
y
dx
4
=
n
4
a
1
e
nx
+
a
2
e
−
nx
+
a
3
cos
nx
+
a
4
sin
nx
=
n
4
y
size 12{ { {d rSup { size 8{4} } y} over { ital "dx" rSup { size 8{4} } } } =n rSup { size 8{4} } left [a rSub { size 8{1} } e rSup { size 8{ ital "nx"} } +a rSub { size 8{2} } e rSup { size 8{ - ital "nx"} } +a rSub { size 8{3} } "cos" ital "nx"+a rSub { size 8{4} } "sin" ital "nx" right ]=n rSup { size 8{4} } y} {}
(8)3. Да се пресмета
dydxdydx size 12{ { { ital "dy"} over { ital "dx"} } } {}, каде y е зададена со параметарските равенки:
x=3at1+t3,y=3at21+t3x=3at1+t3,y=3at21+t3 size 12{x= { {3 ital "at"} over {1+t rSup { size 8{3} } } } ,`````````y= { {3 ital "at" rSup { size 8{2} } } over {1+t rSup { size 8{3} } } } } {}.
Решение.
x
°
=
dx
dt
=
3a
1
+
t
3
−
3
at
3t
2
1
+
t
3
2
=
3a
+
3
at
3
−
9
at
3
1
+
t
3
2
=
3a
−
6
at
3
(
1
+
t
3
)
2
x
°
=
dx
dt
=
3a
1
+
t
3
−
3
at
3t
2
1
+
t
3
2
=
3a
+
3
at
3
−
9
at
3
1
+
t
3
2
=
3a
−
6
at
3
(
1
+
t
3
)
2
size 12{ { size 24{x} } cSup { size 8{ circ } } = { { ital "dx"} over { ital "dt"} } = { {3a left (1+t rSup { size 8{3} } right ) - 3 ital "at"3t rSup { size 8{2} } } over { left (1+t rSup { size 8{3} } right ) rSup { size 8{2} } } } = { {3a+3 ital "at" rSup { size 8{3} } - 9 ital "at" rSup { size 8{3} } } over { left (1+t rSup { size 8{3} } right ) rSup { size 8{2} } } } = { {3a - 6 ital "at" rSup { size 8{3} } } over { \( 1+t rSup { size 8{3} } \) rSup { size 8{2} } } } } {}
(9)
y
°
=
dy
dt
=
6
at
(
1
+
t
3
)
−
3
at
2
3t
2
(
1
+
t
3
)
2
=
6
at
+
6
at
4
−
9
at
4
(
1
+
t
3
)
2
=
6
at
−
3
at
4
(
1
+
t
3
)
y
°
=
dy
dt
=
6
at
(
1
+
t
3
)
−
3
at
2
3t
2
(
1
+
t
3
)
2
=
6
at
+
6
at
4
−
9
at
4
(
1
+
t
3
)
2
=
6
at
−
3
at
4
(
1
+
t
3
)
size 12{ { size 24{y} } cSup { size 8{ circ } } = { { ital "dy"} over { ital "dt"} } = { {6 ital "at" \( 1+t rSup { size 8{3} } \) - 3 ital "at" rSup { size 8{2} } 3t rSup { size 8{2} } } over { \( 1+t rSup { size 8{3} } \) rSup { size 8{2} } } } = { {6 ital "at"+6 ital "at" rSup { size 8{4} } - 9 ital "at" rSup { size 8{4} } } over { \( 1+t rSup { size 8{3} } \) rSup { size 8{2} } } } = { {6 ital "at" - 3 ital "at" rSup { size 8{4} } } over { \( 1+t rSup { size 8{3} } \) } } } {}
(10)dydx=dydtdxdt=6at−3at41+t323a−6at3(1+t3)2=6at−3at43a−6at3=3at2−t33a1−2t3=t2−t31−2t3dydx=dydtdxdt=6at−3at41+t323a−6at3(1+t3)2=6at−3at43a−6at3=3at2−t33a1−2t3=t2−t31−2t3 size 12{ { { ital "dy"} over { ital "dx"} } = { { { { ital "dy"} over { ital "dt"} } } over { { { ital "dx"} over { ital "dt"} } } } = { { { {6 ital "at" - 3 ital "at" rSup { size 8{4} } } over { left (1+t rSup { size 8{3} } right ) rSup { size 8{2} } } } } over { { {3a - 6 ital "at" rSup { size 8{3} } } over { \( 1+t rSup { size 8{3} } \) rSup { size 8{2} } } } } } = { {6 ital "at" - 3 ital "at" rSup { size 8{4} } } over {3a - 6 ital "at" rSup { size 8{3} } } } = { {3 ital "at" left (2 - t rSup { size 8{3} } right )} over {3a left (1 - 2t rSup { size 8{3} } right )} } = { {t left (2 - t rSup { size 8{3} } right )} over {1 - 2t rSup { size 8{3} } } } } {}.
4. Да се докаже дека функцијата зададена со параметарските равенки
x=1+lntt2,y=3+2lnttx=1+lntt2,y=3+2lntt size 12{x= { {1+"ln"t} over {t rSup { size 8{2} } } } `,````y= { {3+2"ln"t} over {t} } } {}, го задоволува равенството:
yy'=2xy'2+1yy'=2xy'2+1 size 12{ ital "yy" rSup { size 8{'} } =2 ital "xy" rSup { size 8{'2} } +1} {}.
Решение.
x
°
=
dx
dt
=
1
t
⋅
t
2
−
1
+
ln
t
2t
t
4
=
t
−
2t
−
2t
ln
t
t
4
=
−
t
−
2t
ln
t
t
4
=
−
t
(
1
+
2
ln
t
)
t
4
=
−
1
+
2
ln
t
t
3
x
°
=
dx
dt
=
1
t
⋅
t
2
−
1
+
ln
t
2t
t
4
=
t
−
2t
−
2t
ln
t
t
4
=
−
t
−
2t
ln
t
t
4
=
−
t
(
1
+
2
ln
t
)
t
4
=
−
1
+
2
ln
t
t
3
size 12{ {x} cSup { size 8{ circ } } = { { ital "dx"} over { ital "dt"} } = { { { {1} over {t} } cdot t rSup { size 8{2} } - left (1+"ln"t right )2t} over {t rSup { size 8{4} } } } = { {t - 2t - 2t"ln"t} over {t rSup { size 8{4} } } } = { { - t - 2t"ln"t} over {t rSup { size 8{4} } } } = - { {t \( 1+2"ln"t \) } over {t rSup { size 8{4} } } } = - { {1+2"ln"t} over {t rSup { size 8{3} } } } } {}
y
°
=
dy
dt
=
2
⋅
1
t
⋅
t
−
(
3
+
2
ln
t
)
⋅
1
t
2
=
2
−
3
−
2
ln
t
t
2
=
−
1
−
2
ln
t
t
2
=
−
1
+
2
ln
t
t
2
y
°
=
dy
dt
=
2
⋅
1
t
⋅
t
−
(
3
+
2
ln
t
)
⋅
1
t
2
=
2
−
3
−
2
ln
t
t
2
=
−
1
−
2
ln
t
t
2
=
−
1
+
2
ln
t
t
2
size 12{ {y} cSup { size 8{ circ } } = { { ital "dy"} over { ital "dt"} } = { {2 cdot { {1} over {t} } cdot t - \( 3+2"ln"t \) cdot 1} over {t rSup { size 8{2} } } } = { {2 - 3 - 2"ln"t} over {t rSup { size 8{2} } } } = { { - 1 - 2"ln"t} over {t rSup { size 8{2} } } } = - { {1+2"ln"t} over {t rSup { size 8{2} } } } } {}
(11)
y
'
=
dy
dx
=
dy
dt
dx
dt
=
−
1
+
2
ln
t
t
2
−
1
+
2
ln
t
t
3
=
t
3
t
2
=
t
y
'
=
dy
dx
=
dy
dt
dx
dt
=
−
1
+
2
ln
t
t
2
−
1
+
2
ln
t
t
3
=
t
3
t
2
=
t
size 12{y rSup { size 8{'} } = { { ital "dy"} over { ital "dx"} } = { { { { ital "dy"} over { ital "dt"} } } over { { { ital "dx"} over { ital "dt"} } } } = { { - { {1+2"ln"t} over {t rSup { size 8{2} } } } } over { - { {1+2"ln"t} over {t rSup { size 8{3} } } } } } = { {t rSup { size 8{3} } } over {t rSup { size 8{2} } } } =t} {}
yy
'
=
3
+
2
ln
t
t
⋅
t
=
3
+
2
ln
t
yy
'
=
3
+
2
ln
t
t
⋅
t
=
3
+
2
ln
t
size 12{ ital "yy" rSup { size 8{'} } = { {3+2"ln"t} over {t} } cdot t=3+2"ln"t} {}
(12)
2
xy
'
2
+
1
=
2
⋅
1
+
ln
t
t
2
⋅
t
2
+
1
=
2
1
+
ln
t
+
1
=
2
+
2
ln
t
+
1
=
3
+
2
ln
t
2
xy
'
2
+
1
=
2
⋅
1
+
ln
t
t
2
⋅
t
2
+
1
=
2
1
+
ln
t
+
1
=
2
+
2
ln
t
+
1
=
3
+
2
ln
t
size 12{2 ital "xy" rSup { size 8{'2} } +1=2 cdot { {1+"ln"t} over {t rSup { size 8{2} } } } cdot t rSup { size 8{2} } +1=2 left (1+"ln"t right )+1=2+2"ln"t+1=3+2"ln"t} {}
(13)Следува
yy'=2xy'2+1yy'=2xy'2+1 size 12{ ital "yy" rSup { size 8{'} } =2 ital "xy" rSup { size 8{'2} } +1} {}.
- Да се докаже дека функцијата
y=y(x)y=y(x) size 12{y=y \( x \) } {}, зададена параметарски:
y=sin2ty=sin2t size 12{y="sin"2t} {},
x=sintx=sint size 12{x="sin"t} {}, ја задоволува релацијата:
(1−x2)y''−xy'+4y=0(1−x2)y''−xy'+4y=0 size 12{ \( 1 - x rSup { size 8{2} } \) y rSup { size 8{"''"} } - ital "xy"'+4y=0} {}.
Решение.
y
°
=
cos
2t
(
2t
)
'
=
2
cos
2t
y
°
=
cos
2t
(
2t
)
'
=
2
cos
2t
size 12{ {y} cSup { size 8{ circ } } ="cos"2t \( 2t \) rSup { size 8{'} } =2"cos"2t} {}
(14)
x
°
=
cos
t
(
t
)
'
=
cos
t
x
°
=
cos
t
(
t
)
'
=
cos
t
size 12{ {x} cSup { size 8{ circ } } ="cos"t \( t \) rSup { size 8{'} } ="cos"t} {}
(15)
y
'
=
2
cos
2t
cos
t
y
'
=
2
cos
2t
cos
t
size 12{y rSup { size 8{'} } = { {2"cos"2t} over {"cos"t} } } {}
(16)
y
°
°
=
−
2
sin
2t
(
2t
)
'
=
−
4
sin
2t
y
°
°
=
−
2
sin
2t
(
2t
)
'
=
−
4
sin
2t
size 12{ {y} cSup { size 8{ circ circ } } = - 2"sin"2t \( 2t \) rSup { size 8{'} } = - 4"sin"2t} {}
(17)
x
°
°
=
−
sin
t
x
°
°
=
−
sin
t
size 12{ {x} cSup { size 8{ circ circ } } = - "sin"t} {}
(18)
y
''
=
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
cos
3
t
y
''
=
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
cos
3
t
size 12{y rSup { size 8{"''"} } = { { - 4"sin"2t"cos"t+2"cos"2t"sin"t} over {"cos" rSup { size 8{3} } t} } } {}
(19)
L
=
(
1
−
sin
2
t
)
⋅
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
cos
3
t
−
sin
t
⋅
2
cos
2t
cos
t
+
4
sin
2t
=
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
−
2
sin
t
cos
2t
+
4
sin
2t
cos
t
cos
t
=
0
cos
t
=
0
=
D
L
=
(
1
−
sin
2
t
)
⋅
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
cos
3
t
−
sin
t
⋅
2
cos
2t
cos
t
+
4
sin
2t
=
−
4
sin
2t
cos
t
+
2
cos
2t
sin
t
−
2
sin
t
cos
2t
+
4
sin
2t
cos
t
cos
t
=
0
cos
t
=
0
=
D
alignl { stack {
size 12{L= \( 1 - "sin" rSup { size 8{2} } t \) cdot { { - 4"sin"2t"cos"t+2"cos"2t"sin"t} over {"cos" rSup { size 8{3} } t} } - "sin"t cdot { {2"cos"2t} over {"cos"t} } +4"sin"2t={}} {} #
= { { - 4"sin"2t"cos"t+2"cos"2t"sin"t - 2"sin"t"cos"2t+4"sin"2t"cos"t} over {"cos"t} } = { {0} over {"cos"t} } =0=D {}
} } {}
(20)- Да се докаже дека функцијата зададена параметарски
y=3t−t3,x=3t2y=3t−t3,x=3t2 size 12{y=3t - t rSup { size 8{3} } ,~x=3t rSup { size 8{2} } } {}, ја задоволува релацијата:
36y'(y−3x)=x+336y'(y−3x)=x+3 size 12{"36"y rSup { size 8{'} } \( y - sqrt {3x} \) =x+3} {}.
Решение.
y
°
=
3
−
3t
2
,
x
°
=
6t
y
°
=
3
−
3t
2
,
x
°
=
6t
size 12{ {y} cSup { size 8{ circ } } =3 - 3t rSup { size 8{2} } ,`````````````````````` {x} cSup { size 8{ circ } } =6t} {}
(21)
y
°
°
=
−
6t
,
x
°
°
=
6
y
°
°
=
−
6t
,
x
°
°
=
6
size 12{ {y} cSup { size 8{ circ circ } } = - 6t`,```````````````````````````` {x} cSup { size 8{ circ circ } } =6`} {}
(22)
y
''
=
y
°
°
x
°
−
x
°
°
y
°
x
°
3
=
−
6t
⋅
6t
−
6
(
3
−
3t
2
)
6
3
⋅
t
3
=
−
36
t
2
−
18
+
18
t
2
6
3
t
3
=
−
18
t
2
−
18
6
3
t
3
=
y
''
=
y
°
°
x
°
−
x
°
°
y
°
x
°
3
=
−
6t
⋅
6t
−
6
(
3
−
3t
2
)
6
3
⋅
t
3
=
−
36
t
2
−
18
+
18
t
2
6
3
t
3
=
−
18
t
2
−
18
6
3
t
3
=
size 12{y rSup { size 8{"''"} } = { { {y} cSup { size 8{ circ circ } } {x} cSup { size 8{ circ } } - {x} cSup { size 8{ circ circ } } {y} cSup { size 8{ circ } } } over { {x} cSup { size 8{ circ } } rSup { size 8{3} } } } = { { - 6t cdot 6t - 6 \( 3 - 3t rSup { size 8{2} } \) } over {6 rSup { size 8{3} } cdot t rSup { size 8{3} } } } = { { - "36"t rSup { size 8{2} } - "18"+"18"t rSup { size 8{2} } } over {6 rSup { size 8{3} } t rSup { size 8{3} } } } = { { - "18"t rSup { size 8{2} } - "18"} over {6 rSup { size 8{3} } t rSup { size 8{3} } } } ={}} {}
(23)
=
−
18
(
t
2
+
1
)
6
3
t
3
=
−
(
t
2
+
1
)
12
t
3
=
−
18
(
t
2
+
1
)
6
3
t
3
=
−
(
t
2
+
1
)
12
t
3
size 12{ {}= { { - "18" \( t rSup { size 8{2} } +1 \) } over {6 rSup { size 8{3} } t rSup { size 8{3} } } } = { { - \( t rSup { size 8{2} } +1 \) } over {"12"t rSup { size 8{3} } } } } {}
(24)
36
y
''
(
y
−
3x
)
=
36
⋅
−
(
t
2
+
1
)
12
t
3
⋅
(
3t
−
t
3
−
3
⋅
3t
2
)
=
=
−
3
t
2
+
1
t
3
(
3t
−
t
3
−
3t
)
=
(
−
3t
2
−
3
)
t
3
(
−
t
3
)
=
−
(
−
3t
2
−
3
)
=
3t
2
+
3
x
+
3
=
3t
2
+
3
36
y
''
(
y
−
3x
)
=
36
⋅
−
(
t
2
+
1
)
12
t
3
⋅
(
3t
−
t
3
−
3
⋅
3t
2
)
=
=
−
3
t
2
+
1
t
3
(
3t
−
t
3
−
3t
)
=
(
−
3t
2
−
3
)
t
3
(
−
t
3
)
=
−
(
−
3t
2
−
3
)
=
3t
2
+
3
x
+
3
=
3t
2
+
3
alignl { stack {
size 12{"36"y rSup { size 8{"''"} } \( y - sqrt {3x} \) ="36" cdot { { - \( t rSup { size 8{2} } +1 \) } over {"12"t rSup { size 8{3} } } } cdot \( 3t - t rSup { size 8{3} } - sqrt {3 cdot 3t rSup { size 8{2} } } \) ={}} {} #
size 12{ {}= { { - 3 left (t rSup { size 8{2} } +1 right )} over {t rSup { size 8{3} } } } \( 3t - t rSup { size 8{3} } - 3t \) = { { \( - 3t rSup { size 8{2} } - 3 \) } over {t rSup { size 8{3} } } } \( - t rSup { size 8{3} } \) = - \( - 3t rSup { size 8{2} } - 3 \) =3t rSup { size 8{2} } +3} {} #
x+3=3t rSup { size 8{2} } +3 {}
} } {}
(25)7. Да се докаже дека функцијата
y=sin(2arcsinx)y=sin(2arcsinx) size 12{y="sin" \( 2"arcsin"x \) } {}ја задоволува релацијата:
(1−x2)y''−xy'+4y=0(1−x2)y''−xy'+4y=0 size 12{ \( 1 - x rSup { size 8{2} } \) y rSup { size 8{"''"} } - ital "xy" rSup { size 8{'} } +4y=0} {}.
Решение.
y
'
=
cos
(
2
arcsin
x
)
⋅
2
⋅
1
1
−
x
2
=
2
⋅
cos
(
2
arcsin
x
)
1
−
x
2
y
'
=
cos
(
2
arcsin
x
)
⋅
2
⋅
1
1
−
x
2
=
2
⋅
cos
(
2
arcsin
x
)
1
−
x
2
size 12{y rSup { size 8{'} } ="cos" \( 2"arcsin"x \) cdot 2 cdot { {1} over { sqrt {1 - x rSup { size 8{2} } } } } =2 cdot { {"cos" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } } {}
(26)
y
''
=
2
⋅
−
sin
(
2
arcsin
x
)
2
1
−
x
2
1
−
x
2
−
cos
(
2
arcsin
x
)
1
⋅
(
−
2x
)
2
1
−
x
2
1
−
x
2
=
y
''
=
2
⋅
−
sin
(
2
arcsin
x
)
2
1
−
x
2
1
−
x
2
−
cos
(
2
arcsin
x
)
1
⋅
(
−
2x
)
2
1
−
x
2
1
−
x
2
=
size 12{y rSup { size 8{"''"} } =2 cdot { { - "sin" \( 2"arcsin"x \) { {2} over { sqrt {1 - x rSup { size 8{2} } } } } sqrt {1 - x rSup { size 8{2} } } - "cos" \( 2"arcsin"x \) { {1 cdot \( - 2x \) } over {2 sqrt {1 - x rSup { size 8{2} } } } } } over {1 - x rSup { size 8{2} } } } ={}} {}
(27)
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
1
−
x
2
1
−
x
2
=
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
1
−
x
2
1
−
x
2
=
size 12{ {}= { { { { - 4 sqrt {1 - x rSup { size 8{2} } } "sin" \( 2"arcsin"x \) +2x"cos" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } } over {1 - x rSup { size 8{2} } } } ={}} {}
(28)
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
(
1
−
x
2
)
1
−
x
2
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
(
1
−
x
2
)
1
−
x
2
size 12{ {}= { { - 4 sqrt {1 - x rSup { size 8{2} } } "sin" \( 2"arcsin"x \) +2x"cos" \( 2"arcsin"x \) } over { \( 1 - x rSup { size 8{2} } \) sqrt {1 - x rSup { size 8{2} } } } } } {}
L
=
(
1
−
x
2
)
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
(
1
−
x
2
)
1
−
x
2
−
L
=
(
1
−
x
2
)
−
4
1
−
x
2
sin
(
2
arcsin
x
)
+
2x
cos
(
2
arcsin
x
)
(
1
−
x
2
)
1
−
x
2
−
size 12{L= \( 1 - x rSup { size 8{2} } \) { { - 4 sqrt {1 - x rSup { size 8{2} } } "sin" \( 2"arcsin"x \) +2x"cos" \( 2"arcsin"x \) } over { \( 1 - x rSup { size 8{2} } \) sqrt {1 - x rSup { size 8{2} } } } } - {}} {}
(29)
−
2
cos
(
2
arcsin
x
)
1
−
x
2
+
4
sin
(
2
arcsin
x
)
=
−
2
cos
(
2
arcsin
x
)
1
−
x
2
+
4
sin
(
2
arcsin
x
)
=
size 12{ - { {2"cos" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } +4"sin" \( 2"arcsin"x \) ={}} {}
(30)
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
1
−
x
2
+
2x
cos
(
2
arcsin
x
)
1
−
x
2
−
=
−
4
1
−
x
2
sin
(
2
arcsin
x
)
1
−
x
2
+
2x
cos
(
2
arcsin
x
)
1
−
x
2
−
size 12{ {}= { { - 4 sqrt {1 - x rSup { size 8{2} } } "sin" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } + { {2x"cos" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } - {}} {}
(31)
−
2
cos
(
2
arcsin
x
)
1
−
x
2
+
4
sin
(
2
arcsin
x
)
=
0
=
D
−
2
cos
(
2
arcsin
x
)
1
−
x
2
+
4
sin
(
2
arcsin
x
)
=
0
=
D
size 12{ - { {2"cos" \( 2"arcsin"x \) } over { sqrt {1 - x rSup { size 8{2} } } } } +4"sin" \( 2"arcsin"x \) =0=D} {}
(32)8. Покажи дека функцијата зададена параметарски:
x
=
2t
+
3t
2
y
=
t
2
+
2t
3
x
=
2t
+
3t
2
y
=
t
2
+
2t
3
alignl { stack {
size 12{x=2t+3t rSup { size 8{2} } } {} #
y=t rSup { size 8{2} } +2t rSup { size 8{3} } {}
} } {}
(33)ја задоволува релацијата
23y=y2y23y=y2y size 12{y= { {y}} sup { ' } rSup { size 8{2} } +2 { {y}} sup { ' } rSup { size 8{3} } } {}.
Решение.
x
°
=
2
+
6t
x
°
=
2
+
6t
size 12{ {x} cSup { size 8{ circ } } =2+6t} {}
(34)
y
°
=
2t
+
6t
2
y
°
=
2t
+
6t
2
size 12{ {y} cSup { size 8{ circ } } =2t+6t"" lSup { size 8{2} } } {}
(35)
y
'
=
y
°
x
°
=
2t
+
6t
2
2
+
6t
=
t
2
+
6t
2
+
6t
=
t
y
'
=
y
°
x
°
=
2t
+
6t
2
2
+
6t
=
t
2
+
6t
2
+
6t
=
t
size 12{ { {y}} sup { ' }= { { {y} cSup { size 8{ circ } } } over { {x} cSup { size 8{ circ } } } } = { {2t+6t rSup { size 8{2} } } over {2+6t} } = { {t left (2+6t right )} over {2+6t} } =t} {}
(36)
t
2
+
2t
3
=
t
2
+
2t
3
t
2
+
2t
3
=
t
2
+
2t
3
size 12{t rSup { size 8{2} } +2t rSup { size 8{3} } =t rSup { size 8{2} } +2t rSup { size 8{3} } } {}
(37)9. Провери дали функцијата зададена параметарски:
x
=
1
+
t
t
3
y
=
3
2t
2
+
2
t
x
=
1
+
t
t
3
y
=
3
2t
2
+
2
t
alignl { stack {
size 12{x= { {1+t} over {t rSup { size 8{3} } } } } {} #
y= { {3} over {2t"" lSup { size 8{2} } } } + { {2} over {t} } {}
} } {}
ја задоволува релацијата
3xy1+y'3xy1+y' size 12{x { {y}} sup { ' } rSup { size 8{3} } =1+ { {y}} sup { ' }} {}.
Решение.
x
°
=
1
+
t
t
3
′
=
t
3
−
1
+
t
⋅
3t
2
t
6
=
t
2
t
−
3
−
3t
t
6
=
−
2t
−
3
t
4
x
°
=
1
+
t
t
3
′
=
t
3
−
1
+
t
⋅
3t
2
t
6
=
t
2
t
−
3
−
3t
t
6
=
−
2t
−
3
t
4
size 12{ {x} cSup { size 8{ circ } } = left ( { {1+t} over {t rSup { size 8{3} } } } right ) rSup { size 8{′} } = { {t rSup { size 8{3} } - left (1+t right ) cdot 3t rSup { size 8{2} } } over {t rSup { size 8{6} } } } = { {t rSup { size 8{2} } left (t - 3 - 3t right )} over {t rSup { size 8{6} } } } = { { - 2t - 3} over {t rSup { size 8{4} } } } } {}
(38)
y
°
=
3
2t
2
+
2
t
′
=
3
+
4t
2t
2
′
=
4
⋅
2t
2
−
3
+
4t
⋅
4t
4t
4
=
8t
2
−
12
t
−
16
t
2
4t
4
=
−
8t
2
−
12
t
4t
4
=
4t
⋅
−
2t
−
3
4t
4
=
−
2t
−
3
t
3
y
°
=
3
2t
2
+
2
t
′
=
3
+
4t
2t
2
′
=
4
⋅
2t
2
−
3
+
4t
⋅
4t
4t
4
=
8t
2
−
12
t
−
16
t
2
4t
4
=
−
8t
2
−
12
t
4t
4
=
4t
⋅
−
2t
−
3
4t
4
=
−
2t
−
3
t
3
alignl { stack {
size 12{ {y} cSup { size 8{ circ } } = left ( { {3} over {2t rSup { size 8{2} } } } + { {2} over {t} } right ) rSup { size 8{′} } = left ( { {3+4t} over {2t rSup { size 8{2} } } } right ) rSup { size 8{′} } = { {4 cdot left (2t rSup { size 8{2} } right ) - left (3+4t right ) cdot 4t} over {4t rSup { size 8{4} } } } = { {8t rSup { size 8{2} } - "12"t - "16"t rSup { size 8{2} } } over {4t rSup { size 8{4} } } } ={}} {} #
= { { - 8t rSup { size 8{2} } - "12"t} over {4t rSup { size 8{4} } } } = { {4t cdot left ( - 2t - 3 right )} over {4t"" lSup { size 8{4} } } } = { { - 2t - 3} over {t rSup { size 8{3} } } } {}
} } {}
(39)
y
'
=
−
2t
−
3
t
3
−
2t
−
3
t
4
=
t
4
t
3
=
t
y
'
=
−
2t
−
3
t
3
−
2t
−
3
t
4
=
t
4
t
3
=
t
size 12{ { {y}} sup { ' }= { { { { - 2t - 3} over {t rSup { size 8{3} } } } } over { { { - 2t - 3} over {t rSup { size 8{4} } } } } } = { {t rSup { size 8{4} } } over {t rSup { size 8{3} } } } =t} {}
3
x
y
1
+
t
t
3
⋅
t
3
=
1
+
t
=
1
+
y
'
3
x
y
1
+
t
t
3
⋅
t
3
=
1
+
t
=
1
+
y
'
size 12{x { {y}} sup { ' } rSup { size 8{3} } = { {1+t} over {t rSup { size 8{3} } } } cdot t rSup { size 8{3} } =1+t=1+ { {y}} sup { ' }} {}
(40)10. Да се покаже дека функцијата
y=x−3x+4y=x−3x+4 size 12{y= { {x - 3} over {x+4} } } {} ја задоволува релацијата
22yy−1y''22yy−1y'' size 12{2 { {y}} sup { ' } rSup { size 8{2} } = left (y - 1 right ) { {y}} sup { '' }} {}.
Решение.
y
'
=
x
−
3
x
+
4
=
x
+
4
−
x
+
3
x
+
4
2
=
7
x
+
4
2
y
'
=
x
−
3
x
+
4
=
x
+
4
−
x
+
3
x
+
4
2
=
7
x
+
4
2
size 12{ { {y}} sup { ' }= left ( { {x - 3} over {x+4} } right )= { {x+4 - x+3} over { left (x+4 right ) rSup { size 8{2} } } } = { {7} over { left (x+4 right ) rSup { size 8{2} } } } } {}
(41)
y
'
'
=
7
x
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4
2
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7
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x
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4
2
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7
⋅
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x
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−
14
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4
x
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4
=
−
14
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+
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3
y
'
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7
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size 12{ { {y}} sup { '' }= left ( { {7} over { left (x+4 right ) rSup { size 8{2} } } } right ) rSup { size 8{′} } = { { left (7 right ) rSup { size 8{′} } left (x+4 right ) rSup { size 8{2} } - 7 cdot 2 left (x+4 right )} over { left (x+4 right ) rSup { size 8{4} } } } = { { - "14" left (x+4 right )} over { left (x+4 right ) rSup { size 8{4} } } } = { { - "14"} over { left (x+4 right ) rSup { size 8{3} } } } } {}
(42)
2
2
y
2
⋅
7
x
+
4
2
2
=
2
⋅
49
x
+
4
4
=
98
x
+
4
4
2
2
y
2
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7
x
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2
2
=
2
⋅
49
x
+
4
4
=
98
x
+
4
4
size 12{2 { {y}} sup { ' } rSup { size 8{2} } =2 cdot left ( { {7} over { left (x+4 right ) rSup { size 8{2} } } } right ) rSup { size 8{2} } =2 cdot { {"49"} over { left (x+4 right ) rSup { size 8{4} } } } = { {"98"} over { left (x+4 right ) rSup { size 8{4} } } } } {}
y
−
1
y
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x
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3
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x
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14
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98
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y
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1
y
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'
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3
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x
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3
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98
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+
4
4
size 12{ left (y - 1 right ) { {y}} sup { '' }= left ( { {x - 3} over {x+4} } - 1 right ) cdot { { - "14"} over { left (x+4 right ) rSup { size 8{3} } } } = { {x - 3 - x - 4} over {x+4} } cdot { { - "14"} over { left (x+4 right ) rSup { size 8{3} } } } = { {"98"} over { left (x+4 right ) rSup { size 8{4} } } } } {}
(43)11. Да се покаже дека функцијата
y=2x−x2y=2x−x2 size 12{y= sqrt {2x - x rSup { size 8{2} } } } {} ја задоволува релацијата
y3y''+1=0y3y''+1=0 size 12{y rSup { size 8{3} } { {y}} sup { '' }+1=0} {}.
Решение.
y
'
=
1
2
2x
−
x
2
(
2x
−
x
2
)
'
=
2
−
2x
2
2x
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x
2
=
2
1
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x
2
2x
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x
2
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1
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x
2x
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y
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2
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2x
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2
)
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2x
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1
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2x
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2x
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x
size 12{ { {y}} sup { ' }= { {1} over {2 sqrt {2x - x rSup { size 8{2} } } } } \( 2x - x rSup { size 8{2} } \) '= { {2 - 2x} over {2 sqrt {2x - x rSup { size 8{2} } } } } = { {2 left (1 - x right )} over {2 sqrt {2x - x rSup { size 8{2} } } } } = { {1 - x} over { sqrt {2x - x} } } } {}
(44)
y
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x
2x
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x
2
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=
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1
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2
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2x
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3
y
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2x
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2x
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1
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x
2
2x
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1
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2x
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2
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2x
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x
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1
2x
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2
3
alignl { stack {
size 12{ { {y}} sup { '' }= left ( { {1 - x} over { sqrt {2x - x rSup { size 8{2} } } } } right ) rSup { size 8{'} } = { { - sqrt {2x - x rSup { size 8{2} } } - left (1 - x right ) cdot { {1} over {2 sqrt {2x - x rSup { size 8{2} } } } } cdot \( 2 - 2x \) } over {2x - x rSup { size 8{2} } } } = { { { { - 2x+x rSup { size 8{2} } - left (1 - x right ) rSup { size 8{2} } } over { sqrt {2x - x rSup { size 8{2} } } } } } over { { {2x - x rSup { size 8{2} } } over {1} } } } ={}} {} #
= { { - 2x+x rSup { size 8{2} } - 1+2x - x rSup { size 8{2} } } over { left (2x - x rSup { size 8{2} } right ) sqrt {2x - x rSup { size 8{2} } } } } = { { - 1} over { left ( sqrt {2x - x rSup { size 8{2} } } right ) rSup { size 8{3} } } } {}
} } {}
(45)
y
3
y
'
'
+
1
=
2x
−
x
2
3
⋅
−
1
2x
−
x
2
3
+
1
=
(
−
1
)
+
1
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y
3
y
'
'
+
1
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2x
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x
2
3
⋅
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1
2x
−
x
2
3
+
1
=
(
−
1
)
+
1
=
0
size 12{y rSup { size 8{3} } { {y}} sup { '' }+1= left ( sqrt {2x - x rSup { size 8{2} } } right ) rSup { size 8{3} } cdot { { - 1} over { left ( sqrt {2x - x rSup { size 8{2} } } right ) rSup { size 8{3} } } } +1= \( - 1 \) +1=0} {}
(46)12. Да се покаже дека функцијата
y=e4x+2e−xy=e4x+2e−x size 12{y=e rSup { size 8{4x} } +2e rSup { size 8{ - x} } } {} ја задоволува релацијата
y'''−13 {y'−12y=0y'''−13 {y'−12y=0 size 12{ { {y}} sup { ''' } - "13 {" ital {y}} sup { ' } - "12"y=0} {}.
Решение.
y
'
=
e
4x
+
2
e
−
x
=
e
4x
⋅
4x
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+
2
e
−
x
⋅
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x
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4
e
4x
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2
e
−
x
y
'
=
e
4x
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2
e
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x
=
e
4x
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2
e
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x
⋅
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e
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size 12{ { {y}} sup { ' }=e rSup { size 8{4x} } +2e rSup { size 8{ - x} } =e rSup { size 8{4x} } cdot left (4x right ) rSup { size 8{′} } +2e rSup { size 8{ - x} } cdot left ( - x right ) rSup { size 8{′} } =4e rSup { size 8{4x} } - 2e rSup { size 8{ - x} } } {}
(47)
y
'
'
=
4
e
4x
⋅
4x
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−
2
e
−
x
⋅
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x
′
=
16
e
4x
+
2
e
−
x
y
'
'
=
4
e
4x
⋅
4x
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−
2
e
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x
⋅
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x
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=
16
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4x
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2
e
−
x
size 12{ { {y}} sup { '' }=4e rSup { size 8{4x} } cdot left (4x right ) rSup { size 8{′} } - 2e rSup { size 8{ - x} } cdot left ( - x right ) rSup { size 8{′} } ="16"e rSup { size 8{4x} } +2e rSup { size 8{ - x} } } {}
(48)
y
'
'
'
=
16
e
4x
⋅
4x
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+
2
e
−
x
⋅
−
x
′
=
64
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y
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'
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16
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=
64
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size 12{ { {y}} sup { ''' }="16"e rSup { size 8{4x} } cdot left (4x right ) rSup { size 8{′} } +2e rSup { size 8{ - x} } cdot left ( - x right ) rSup { size 8{′} } ="64"e rSup { size 8{4x} } - 2e rSup { size 8{ - x} } } {}
(49)
y
'
'
'
−
13 {
y
'
−
12
y
=
64
e
4x
−
2
e
−
x
−
13
4
e
4x
−
2
e
−
x
−
12
e
4x
+
2
e
−
x
=
64
e
4x
−
2
e
−
x
−
52
e
4x
+
26
e
−
x
−
12
e
4x
−
24
e
−
x
=
0
y
'
'
'
−
13 {
y
'
−
12
y
=
64
e
4x
−
2
e
−
x
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13
4
e
4x
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2
e
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2
e
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64
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52
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4x
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26
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e
4x
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24
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x
=
0
alignl { stack {
size 12{ { {y}} sup { ''' } - "13 {" ital {y}} sup { ' } - "12"y="64"e rSup { size 8{4x} } - 2e rSup { size 8{ - x} } - "13" left (4e rSup { size 8{4x} } - 2e rSup { size 8{ - x} } right ) - "12" left (e rSup { size 8{4x} } +2e rSup { size 8{ - x} } right )={}} {} #
"64"e rSup { size 8{4x} } - 2e rSup { size 8{ - x} } - "52"e rSup { size 8{4x} } +"26"e rSup { size 8{ - x} } - "12"e rSup { size 8{4x} } - "24"e rSup { size 8{ - x} } =0 {}
} } {}
13. Да се покаже дека функцијата
y=ex+e−xy=ex+e−x size 12{y=e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } {} ја задоволува релацијата
xy''+12y'−14y=0xy''+12y'−14y=0 size 12{x { {y}} sup { '' }+ { {1} over {2} } { {y}} sup { ' } - { {1} over {4} } y=0} {}.
Решение.
{}
y
'
=
e
x
+
e
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x
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=
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x
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2
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+
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y
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alignl { stack {
size 12{ { {y}} sup { ' }= left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) rSup { size 8{′} } =e rSup { size 8{ sqrt {x} } } cdot { {1} over {2 sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } cdot left ( - { {1} over {2 sqrt {x} } } right )= { {e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } } over {2 sqrt {x} } } } {} #
{}
} } {}
y
'
'
=
e
x
−
e
−
x
2
x
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=
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x
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4x
=
y
'
'
=
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4x
=
size 12{ { {y}} sup { '' }= left ( { {e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } } over {2 sqrt {x} } } right ) rSup { size 8{′} } = { { left (e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - nroot {} {x} } } right ) rSup { size 12{′}} cdot 2 sqrt {x} - left (e rSup { sqrt {x} } size 12{ - e rSup { - sqrt {x} } } right ) size 12{ cdot 2 cdot left ( sqrt {x} right ) rSup { size 12{′}} }} over {4x} } ={}} {}
(50)
=
e
x
1
2
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4x
=
=
e
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⋅
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size 12{ {}= { { left [e rSup { size 8{ sqrt {x} } } { {1} over {2 sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } cdot left ( - { {1} over {2 sqrt {x} } } right ) right ] cdot 2 sqrt {x} - left (e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } right ) cdot 2 cdot { {1} over {2 sqrt {x} } } } over {4x} } ={}} {}
(51)
=
e
x
+
e
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x
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alignl { stack {
size 12{ {}= { { { {e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over {2 sqrt {x} } } cdot 2 sqrt {x} - { {e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } } over { sqrt {x} } } } over {4x} } = { { { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} - e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over { sqrt {x} } } } over { { {4x} over {1} } } } ={}} {} #
= { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} - e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over {4x sqrt {x} } } {}
} } {}
(52)
x
y
'
'
+
1
2
y
'
−
1
4
y
=
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+
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4x
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4
e
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x
y
'
'
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1
2
y
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4
y
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=
size 12{x { {y}} sup { '' }+ { {1} over {2} } { {y}} sup { ' } - { {1} over {4} } y=x cdot { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} - e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over {4x sqrt {x} } } + { {1} over {2} } cdot { {e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } } over {2 sqrt {x} } } - { {1} over {4} } left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right )={}} {}
(53)
=
e
x
+
e
−
x
⋅
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−
e
x
+
e
−
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+
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x
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=
=
e
x
+
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x
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+
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+
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x
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4
x
−
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x
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4
=
size 12{ {}= { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} - e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } +e rSup { size 8{ sqrt {x} } } - e rSup { size 8{ - sqrt {x} } } } over {4 sqrt {x} } } - { {e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over {4} } ={}} {}
(54)
=
e
x
+
e
−
x
⋅
x
4
x
−
e
x
+
e
−
x
4
=
e
x
+
e
−
x
⋅
x
−
e
x
+
e
−
x
⋅
x
4
x
=
0
=
e
x
+
e
−
x
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4
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x
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x
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x
=
0
size 12{ {}= { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} } over {4 sqrt {x} } } - { {e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } } over {4} } = { { left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} - left (e rSup { size 8{ sqrt {x} } } +e rSup { size 8{ - sqrt {x} } } right ) cdot sqrt {x} } over {4 sqrt {x} } } =0} {}
(55)14. Да се покаже дека функцијата
y=eaarcsinxy=eaarcsinx size 12{y=e rSup { size 8{a"arcsin"x} } } {} ја задоволува релацијата
1−x2y''−xy'−a2y=01−x2y''−xy'−a2y=0 size 12{ left (1 - x"" lSup { size 8{2} } right ) { {y}} sup { '' } - x { {y}} sup { ' } - a rSup { size 8{2} } y=0} {}.
Решение.
y
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e
a
arcsin
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e
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arcsin
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arcsin
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ae
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arcsin
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y
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arcsin
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size 12{ { {y}} sup { ' }= left (e rSup { size 8{a"arcsin"x} } right ) rSup { size 8{′} } =e rSup { size 8{a"arcsin"x} } a cdot left ("arcsin"x right ) rSup { size 8{′} } = ital "ae" rSup { size 8{a"arcsin"x} } { {1} over { sqrt {1 - x rSup { size 8{2} } } } } = { { ital "ae" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } } {}
(56)
y
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ae
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arcsin
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1
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x
2
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ae
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arcsin
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ae
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arcsin
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2x
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y
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arcsin
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arcsin
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ae
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size 12{ { {y}} sup { '' }= left ( { { ital "ae" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } right ) rSup { size 8{′} } = { {a { { ital "ae" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } cdot sqrt {1 - x rSup { size 8{2} } } - ital "ae" rSup { size 8{a"arcsin"x} } { { - 2x} over {2 sqrt {1 - x rSup { size 8{2} } } } } } over {1 - x rSup { size 8{2} } } } ={}} {}
(57)
=
a
2
e
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arcsin
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axe
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arcsin
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a
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arcsin
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axe
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arcsin
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a
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axe
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a
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arcsin
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axe
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arcsin
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size 12{ {}= { { { {a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } sqrt {1 - x rSup { size 8{2} } } + ital "axe" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } } over { { {1 - x rSup { size 8{2} } } over {1} } } } = { {a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } sqrt {1 - x rSup { size 8{2} } } + ital "axe" rSup { size 8{a"arcsin"x} } } over { left (1 - x rSup { size 8{2} } right ) sqrt {1 - x rSup { size 8{2} } } } } } {}
(58)
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size 12{ left (1 - x rSup { size 8{2} } right ) { {y}} sup { '' } - x { {y}} sup { ' } - a rSup { size 8{2} } y= left (1 - x rSup { size 8{2} } right ) { {a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } sqrt {1 - x rSup { size 8{2} } } + ital "axe" rSup { size 8{a"arcsin"x} } } over { left (1 - x rSup { size 8{2} } right ) sqrt {1 - x rSup { size 8{2} } } } } - x { { ital "ae" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } - a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } ={}} {}
(59)
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size 12{ {}= { {a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } sqrt {1 - x rSup { size 8{2} } } + ital "axe" rSup { size 8{a"arcsin"x} } - ital "axe" rSup { size 8{a"arcsin"x} } } over { sqrt {1 - x rSup { size 8{2} } } } } - a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } ={}} {}
(60)
=
a
2
e
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arcsin
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1
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arcsin
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0
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0
size 12{ {}= { {a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } sqrt {1 - x rSup { size 8{2} } } } over { sqrt {1 - x rSup { size 8{2} } } } } - a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } =a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } - a rSup { size 8{2} } e rSup { size 8{a"arcsin"x} } =0} {}
(61)15. Да се провери дека функцијата
y=x+x2+1ky=x+x2+1k size 12{y= left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } {} ја задоволува релацијата
1+x2y''+xy'−k2y=01+x2y''+xy'−k2y=0 size 12{ left (1+x rSup { size 8{2} } right ) { {y}} sup { '' }+x { {y}} sup { ' } - k rSup { size 8{2} } y=0} {}.
Решение.
y
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k
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1
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size 12{ { {y}} sup { ' }=k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k - 1} } cdot left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{′} } ={}} {}
(62)
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size 12{ {}=k left (x+ sqrt {x rSup { size 8{x} } +1} right ) rSup { size 8{k - 1} } cdot left (1+ { {2x} over {2 sqrt {x rSup { size 8{2} } +1} } } right )=k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k - 1} } cdot { { sqrt {x rSup { size 8{2} } +1} +x} over { sqrt {x rSup { size 8{2} } +1} } } ={}} {}
(63)=kx+x2+1kx+x2+1⋅x2+1+xx2+1=kx+x2+1kx2+1=kx+x2+1kx+x2+1⋅x2+1+xx2+1=kx+x2+1kx2+1 size 12{ {}=k { { left (x+ sqrt {x rSup { size 8{2} } +1} right )"" lSup { size 8{k} } } over { left (x+ sqrt {x rSup { size 8{2} } +1} right )} } cdot { { left ( sqrt {x rSup { size 8{2} } +1} +x right )} over { sqrt {x rSup { size 8{2} } +1} } } = { {k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } } {}.
y
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y
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size 12{ { {y}} sup { '' }= { {k cdot k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k - 1} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{′} } cdot sqrt {x rSup { size 8{2} } +1} - k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } cdot { {2x} over {2 sqrt {x rSup { size 8{2} } +1} } } } over {x rSup { size 8{2} } +1} } ={}} {}
=
k
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size 12{ { { {}= { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right )} over { left (x+ sqrt {x rSup { size 8{x} } +1} right )} } rSup { size 8{k} } cdot left (1+ { {2x} over {2 sqrt {x rSup { size 8{2} } +1} } } right ) cdot sqrt {x rSup { size 8{2} } +1} - { { ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } } over {x rSup { size 8{2} } +1} } ={}} {}
(64)
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size 12{ {}= { { { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } left ( sqrt {x rSup { size 8{2} } +1} right ) left ( sqrt {x rSup { size 8{2} } +1} +x right )} over { left (x+ sqrt {x rSup { size 8{2} } +1} right ) sqrt {x rSup { size 8{2} } +1} } } - { { ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } } over {x rSup { size 8{2} } +1} } ={}} {}
(65)
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size 12{ {}= { { { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } sqrt {x rSup { size 8{2} } +1} - ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } } over { { {x"" lSup { size 8{2} } +1} over {1} } } } ={}} {}
(66)
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size 12{ {}= { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } sqrt {x rSup { size 8{2} } +1} - ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { left (x rSup { size 8{2} } +1 right ) sqrt {x rSup { size 8{2} } +1} } } } {}
(67)
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size 12{ left (1+x rSup { size 8{2} } right ) { {y}} sup { '' }+x { {y}} sup { ' } - k rSup { size 8{2} } y={}} {}
(68)
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alignl { stack {
size 12{ {}= left (1+x rSup { size 8{2} } right ) cdot { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } sqrt {x rSup { size 8{2} } +1} - ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right )} over { left (x rSup { size 8{2} } +1 right ) sqrt {x rSup { size 8{2} } +1} } } +{}} {} #
+x { {k left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } - k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } ={} {}
} } {}
(69)
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size 12{ {}= { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } sqrt {x rSup { size 8{2} } +1} - ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } + ital "kx" left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } } over { sqrt {x rSup { size 8{2} } +1} } } - k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } ={}} {}
(70)
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size 12{ {}= { {k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } sqrt {x rSup { size 8{2} } +1} } over { sqrt {x rSup { size 8{2} } +1} } } - k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } ={}} {}
(71)=k2x+x2+1k−k2x+x2+1k=0=k2x+x2+1k−k2x+x2+1k=0 size 12{ {}=k rSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } - k"" lSup { size 8{2} } left (x+ sqrt {x rSup { size 8{2} } +1} right ) rSup { size 8{k} } =0} {}.
16. Да се покаже дека функцијата
y=exsinxy=exsinx size 12{y=e rSup { size 8{x} } "sin"x} {} ја задоволува релацијата
y''−2y'+2y=0y''−2y'+2y=0 size 12{y"''" - 2y'+2y=0} {}.
Решение.
y
'
=
e
x
'sin
x
+
e
x
sin
x
'
=
e
x
sin
x
+
e
x
cos
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y
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=
e
x
'sin
x
+
e
x
sin
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=
e
x
sin
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e
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cos
x
size 12{y'= left (e rSup { size 8{x} } right )"'sin"x+e rSup { size 8{x} } left ("sin"x right )'=e rSup { size 8{x} } "sin"x+e rSup { size 8{x} } "cos"x} {}
(72)
y
''
=
e
x
'sin
x
+
e
x
sin
x
'
+
e
x
'cos
x
+
e
x
cos
x
'
=
y
''
=
e
x
'sin
x
+
e
x
sin
x
'
+
e
x
'cos
x
+
e
x
cos
x
'
=
size 12{y"''"= left (e rSup { size 8{x} } right )"'sin"x+e rSup { size 8{x} } left ("sin"x right )'+ left (e rSup { size 8{x} } right )"'cos"x+e rSup { size 8{x} } left ("cos"x right )'={}} {}
(73)
=
e
x
sin
x
+
e
x
cos
x
+
e
x
cos
x
−
e
x
sin
x
=
2
e
x
cos
x
=
e
x
sin
x
+
e
x
cos
x
+
e
x
cos
x
−
e
x
sin
x
=
2
e
x
cos
x
alignl { stack {
size 12{ {}=e rSup { size 8{x} } "sin"x+e rSup { size 8{x} } "cos"x+e rSup { size 8{x} } "cos"x - e rSup { size 8{x} } "sin"x={}} {} #
=2e rSup { size 8{x} } "cos"x {}
} } {}
(74)
y
''
−
2y
'
+
2y
=
2
e
x
cos
x
−
2
e
x
sin
x
+
e
x
cos
x
+
2
e
x
sin
x
=
y
''
−
2y
'
+
2y
=
2
e
x
cos
x
−
2
e
x
sin
x
+
e
x
cos
x
+
2
e
x
sin
x
=
size 12{y"''" - 2y'+2y=2e rSup { size 8{x} } "cos"x - 2 left (e rSup { size 8{x} } "sin"x+e rSup { size 8{x} } "cos"x right )+2e rSup { size 8{x} } "sin"x={}} {}
(75)=2excosx−2exsinx−2excosx+2exsinx=0=2excosx−2exsinx−2excosx+2exsinx=0 size 12{ {}=2e rSup { size 8{x} } "cos"x - 2e rSup { size 8{x} } "sin"x - 2e rSup { size 8{x} } "cos"x+2e rSup { size 8{x} } "sin"x=0} {}.
