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Задачи од изводи и релации

Module by: Beti Andonovic. E-mail the author

Summary: Solved exercises on derivatives and relations Решени задачи од изводи и релации

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,woconst)(A,B,w,woconst) 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+a2enx+a3cosnx+a4sinnxa1enx+a2enx+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,nconst)(a1,a2,a3,a4,nconst) 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=6at3at41+t323a6at3(1+t3)2=6at3at43a6at3=3at2t33a12t3=t2t312t3dydx=dydtdxdt=6at3at41+t323a6at3(1+t3)2=6at3at43a6at3=3at2t33a12t3=t2t312t3 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} {}.

  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: (1x2)y''xy'+4y=0(1x2)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)

  1. Да се докаже дека функцијата зададена параметарски y=3tt3,x=3t2y=3tt3,x=3t2 size 12{y=3t - t rSup { size 8{3} } ,~x=3t rSup { size 8{2} } } {}, ја задоволува релацијата: 36y'(y3x)=x+336y'(y3x)=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: (1x2)y''xy'+4y=0(1x2)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=x3x+4y=x3x+4 size 12{y= { {x - 3} over {x+4} } } {} ја задоволува релацијата 22yy1y''22yy1y'' 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 + 4 2 = 7 x + 4 2 7 2 x + 4 x + 4 4 = 14 x + 4 x + 4 4 = 14 x + 4 3 y ' ' = 7 x + 4 2 = 7 x + 4 2 7 2 x + 4 x + 4 4 = 14 x + 4 x + 4 4 = 14 x + 4 3 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 7 x + 4 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 ' ' = x 3 x + 4 1 14 x + 4 3 = x 3 x 4 x + 4 14 x + 4 3 = 98 x + 4 4 y 1 y ' ' = x 3 x + 4 1 14 x + 4 3 = x 3 x 4 x + 4 14 x + 4 3 = 98 x + 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=2xx2y=2xx2 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 x 2 = 2 1 x 2 2x x 2 = 1 x 2x x y ' = 1 2 2x x 2 ( 2x x 2 ) ' = 2 2x 2 2x x 2 = 2 1 x 2 2x x 2 = 1 x 2x 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 ' ' = 1 x 2x x 2 ' = 2x x 2 1 x 1 2 2x x 2 ( 2 2x ) 2x x 2 = 2x + x 2 1 x 2 2x x 2 2x x 2 1 = 2x + x 2 1 + 2x x 2 2x x 2 2x x 2 = 1 2x x 2 3 y ' ' = 1 x 2x x 2 ' = 2x x 2 1 x 1 2 2x x 2 ( 2 2x ) 2x x 2 = 2x + x 2 1 x 2 2x x 2 2x x 2 1 = 2x + x 2 1 + 2x x 2 2x x 2 2x x 2 = 1 2x x 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 = 0 y 3 y ' ' + 1 = 2x x 2 3 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+2exy=e4x+2ex 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 + 2 e x x = 4 e 4x 2 e x y ' = e 4x + 2 e x = e 4x 4x + 2 e x x = 4 e 4x 2 e x 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 2 e x x = 16 e 4x + 2 e x y ' ' = 4 e 4x 4x 2 e x x = 16 e 4x + 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 + 2 e x x = 64 e 4x 2 e x y ' ' ' = 16 e 4x 4x + 2 e x x = 64 e 4x 2 e x 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 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 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+exy=ex+ex 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 x = e x 1 2 x + e x 1 2 x = e x e x 2 x y ' = e x + e x = e x 1 2 x + e x 1 2 x = e x e x 2 x 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 = e x e x 2 x e x e x 2 x 4x = y ' ' = e x e x 2 x = e x e x 2 x e x e x 2 x 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 x e x 1 2 x 2 x e x e x 2 1 2 x 4x = = e x 1 2 x e x 1 2 x 2 x e x e x 2 1 2 x 4x = 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 x 2 x 2 x e x e x x 4x = e x + e x x e x + e x x 4x 1 = e x + e x x e x + e x 4x x = e x + e x 2 x 2 x e x e x x 4x = e x + e x x e x + e x x 4x 1 = e x + e x x e x + e x 4x x 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 = x e x + e x x e x + e x 4x x + 1 2 e x e x 2 x 1 4 e x + e x = x y ' ' + 1 2 y ' 1 4 y = x e x + e x x e x + e x 4x x + 1 2 e x e x 2 x 1 4 e x + e x = 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 x e x + e x + e x e x 4 x e x + e x 4 = = e x + e x x e x + e x + e x e x 4 x e x + e x 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 x 4 x e x + e x 4 = e x + e x x e x + e x x 4 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 1x2y''xy'a2y=01x2y''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 ' = e a arcsin x = e a arcsin x a arcsin x = ae a arcsin x 1 1 x 2 = ae a arcsin x 1 x 2 y ' = e a arcsin x = e a arcsin x a arcsin x = ae a arcsin x 1 1 x 2 = ae a arcsin x 1 x 2 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 ' ' = ae a arcsin x 1 x 2 = a ae a arcsin x 1 x 2 1 x 2 ae a arcsin x 2x 2 1 x 2 1 x 2 = y ' ' = ae a arcsin x 1 x 2 = a ae a arcsin x 1 x 2 1 x 2 ae a arcsin x 2x 2 1 x 2 1 x 2 = 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 a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 1 = a 2 e a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 = a 2 e a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 1 = a 2 e a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 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)
1 x 2 y ' ' x y ' a 2 y = 1 x 2 a 2 e a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 x ae a arcsin x 1 x 2 a 2 e a arcsin x = 1 x 2 y ' ' x y ' a 2 y = 1 x 2 a 2 e a arcsin x 1 x 2 + axe a arcsin x 1 x 2 1 x 2 x ae a arcsin x 1 x 2 a 2 e a arcsin x = 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)
= a 2 e a arcsin x 1 x 2 + axe a arcsin x axe a arcsin x 1 x 2 a 2 e a arcsin x = = a 2 e a arcsin x 1 x 2 + axe a arcsin x axe a arcsin x 1 x 2 a 2 e a arcsin x = 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 a arcsin x 1 x 2 1 x 2 a 2 e a arcsin x = a 2 e a arcsin x a 2 e a arcsin x = 0 = a 2 e a arcsin x 1 x 2 1 x 2 a 2 e a arcsin x = a 2 e a arcsin x a 2 e a arcsin x = 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 ' = k x + x 2 + 1 k 1 x + x 2 + 1 = y ' = k x + x 2 + 1 k 1 x + x 2 + 1 = 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)
= k x + x x + 1 k 1 1 + 2x 2 x 2 + 1 = k x + x 2 + 1 k 1 x 2 + 1 + x x 2 + 1 = = k x + x x + 1 k 1 1 + 2x 2 x 2 + 1 = k x + x 2 + 1 k 1 x 2 + 1 + x x 2 + 1 = 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+1x2+1+xx2+1=kx+x2+1kx2+1=kx+x2+1kx+x2+1x2+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 ' ' = k k x + x 2 + 1 k 1 x + x 2 + 1 x 2 + 1 k x + x 2 + 1 k 2x 2 x 2 + 1 x 2 + 1 = y ' ' = k k x + x 2 + 1 k 1 x + x 2 + 1 x 2 + 1 k x + x 2 + 1 k 2x 2 x 2 + 1 x 2 + 1 = 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 2 x + x 2 + 1 x + x x + 1 k 1 + 2x 2 x 2 + 1 x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 = = k 2 x + x 2 + 1 x + x x + 1 k 1 + 2x 2 x 2 + 1 x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 = 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)
= k 2 x + x 2 + 1 k x 2 + 1 x 2 + 1 + x x + x 2 + 1 x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 = = k 2 x + x 2 + 1 k x 2 + 1 x 2 + 1 + x x + x 2 + 1 x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 = 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)
= k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 1 = = k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 1 = 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)
= k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 = k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k x 2 + 1 x 2 + 1 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)
1 + x 2 y ' ' + x y ' k 2 y = 1 + x 2 y ' ' + x y ' k 2 y = size 12{ left (1+x rSup { size 8{2} } right ) { {y}} sup { '' }+x { {y}} sup { ' } - k rSup { size 8{2} } y={}} {}
(68)
= 1 + x 2 k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 x 2 + 1 x 2 + 1 + + x k x + x 2 + 1 k x 2 + 1 k 2 x + x 2 + 1 k = = 1 + x 2 k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 x 2 + 1 x 2 + 1 + + x k x + x 2 + 1 k x 2 + 1 k 2 x + x 2 + 1 k = 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)
= k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k + kx x + x 2 + 1 k x 2 + 1 k 2 x + x 2 + 1 k = = k 2 x + x 2 + 1 k x 2 + 1 kx x + x 2 + 1 k + kx x + x 2 + 1 k x 2 + 1 k 2 x + x 2 + 1 k = 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)
= k 2 x + x 2 + 1 k x 2 + 1 x 2 + 1 k 2 x + x 2 + 1 k = = k 2 x + x 2 + 1 k x 2 + 1 x 2 + 1 k 2 x + x 2 + 1 k = 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+1kk2x+x2+1k=0=k2x+x2+1kk2x+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 x y ' = e x 'sin x + e x sin x ' = e x sin 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 )'=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)

=2excosx2exsinx2excosx+2exsinx=0=2excosx2exsinx2excosx+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} {}.

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