Tentukan solusi dari persamaan berikut!!!
y(1 – x)dx + x2(1 – y)dy = 0
Jawab:
y(1-x)dx+ x^2 (1-y)dy=0
x^2 (1-y)dy=- y(1-x)dx Kedua ruas di kali 1/(x^2 y)
((1-y))/y dy = - ((1-x))/x^2 dx
∫▒〖((1-y))/y dy〗 = -∫▒〖((1-x))/x^2 dx〗
∫▒〖1/y dy-〗 ∫▒〖y/y dy〗 = -(∫▒〖1/x^2 dx〗-∫▒〖x/x^2 dx〗)
∫▒〖1/y dy-〗 ∫▒〖y/y dy〗 = -∫▒〖1/x^2 dx〗+∫▒〖x/x^2 dx〗
∫▒〖1/y dy-〗 ∫▒1dy = -∫▒〖x^(-2) dx〗+∫▒〖1/x dx〗
ln〖(y) 〗– ln(x) = x^(-1)+ y+c_3+ c_4-c_2-c_1
ln〖((y))/((x)) 〗= 1/x +y+C
2xy(4 – y2)dx + (y – 1)(x2 +2)dy = 0
Jawab:
2xy(4- y^2 )dx+(y-1)(x^2+ 2)dy=0
(y-1)(x^2+ 2)dy= -2xy(4- y^2 )dx
(y-1)(x^2+ 2)dy= -2x(4y- y^3 )dx Kedua Ruas di kali 1/((4y - y^3 )(x^2+ 2))
((y - 1))/((4y - y^3)) dy = - 2x/(x^2+ 2) dx
((y - 1))/(y(4 - y^2)) dy = - 2x/(x^2+ 2) dx
∫▒〖((y – 1))/y(4 – y^2 ) dy〗 = - ∫▒〖2x/(x^2+ 2) dx〗
∫▒〖y/(y(4 - y^2)) dy - 〗 ∫▒〖1/(y(4 - y^2)) dy=〗 - ∫▒〖2x/(x^2+ 2) dx〗
∫▒〖1/((4 - y^2)) dy - 〗 ∫▒〖1/(y(4 - y^2)) dy=〗 - ∫▒〖2x/(x^2+ 2) dx〗
Selesaikan satu persatu...
∫▒〖2x/(x^2+ 2) dx〗=⋯
Misal:
x^2+ 2=u
2xdx=du
Jadi,
∫▒〖2x/(x^2+ 2) dx〗= ∫▒〖du/u=〗 ln(u)+ c=ln〖(x^2 〗+ 2)+c
∫▒〖1/((4 – y^2 ) ) dy= 〗 1/4 ln〖(2+y)/(2-y)〗+ c
∫▒〖1/(y(4 - y^2)) dy=⋯〗
Misalkan:
1/(y(4 - y^2))= A/y+ (By+C )/((4 - y^2)) = (A(4-y^2 )+y(By+C) )/(y(4 - y^2))
Dari persamaan: A(4-y^2 )+y(By+C)=1,diperoleh:
A(4-y^2 )+y(By+C)=1
4A-Ay^2+ By^2+ Cy=1
4A-(A- B)y^2+ Cy=1
4A=1
A= 1/4
A - B=0
1/4-B=0
B=1/4
C=0
Dengan demikian diperoleh nilai A= 1/4,B= 1/4,dan C=0 .
Subtitusi nilai-nilai tersebut kedalam persamaan sebelumya kemudian integralkan.
1/(y(4 - y^2))= (1/4)/y+ (1/4 y)/((4 - y^2))
∫▒〖1/(y(4 - y^2)) dy 〗=∫▒(1/4)/y dy+∫▒〖(1/4 y )/((4 - y^2)) dy〗
∫▒〖1/(y(4 - y^2)) dy 〗=1/4 ∫▒1/y dy+1/4 ∫▒〖(y )/((4 - y^2)) dy〗
∫▒〖1/(y(4 - y^2)) dy 〗=1/4 ln(y)+ 1/4 x 1/4 ln (2 + y)/(2 - y)+ c
∫▒〖1/(y(4 - y^2)) dy 〗=1/4 ln(y)+ 1/8 ln (2 + y)/(2 - y)+ c
Dengan demikian :
∫▒〖1/((4 - y^2)) dy - 〗 ∫▒〖1/(y(4 - y^2)) dy=〗 - ∫▒〖2x/(x^2+ 2) dx〗
1/4 ln〖(2+y)/(2-y)〗+ c_1-(1/4 ln(y)+ 1/8 ln (2 + y)/(2 - y)+ c_2)= -(ln〖(x^2 〗+ 2)+c_3)
2 ln〖(2+y)/(2-y)〗+ c_1-2ln(y)- ln (2 + y)/(2 - y)- 8c_2= -8ln〖(x^2 〗+ 2)- 8c_3
2 ln〖(2+y)/(2-y)〗-2 ln(y)- ln (2 + y)/(2 - y)+8 ln〖(x^2 〗+ 2)= - 〖8c〗_3-8 c_1+8 c_2
ln (2 + y)/(2 - y)- 2 ln(y)+8 ln〖(x^2 〗+ 2)=C
(1 + 2x2)y dy/dx = 2x(1 + y2)
Jawab:
(1+2x^2 )y dy/dx=2x(1+ y^2 )
(1+2x^2 )y dy=2x(1+ y^2 )dx Kedua ruas di kali 1/((1+2x^2 )(1+ y^2 ))
y/((1+ y^2 )) dy=2x/((1+2 x^2 )) dx
∫▒y/((1+ y^2 )) dy=∫▒2x/((1+2 x^2 )) dx
Selesaikan satu persatu...
∫▒2x/((1+2 x^2 )) dx=⋯
Misal:
(1+2x^2 )=u
(4x)dx=du
(2x)dx=1/2 du
Jadi,
∫▒2x/((1+2 x^2 )) dx= ∫▒(1/2 du)/u= 1/2 ∫▒du/u=1/2 ln(u)+ c= 1/2 ln(1+2x^2 )+ c
∫▒y/((1+ y^2 )) dy=⋯
Misal:
1+ y^2=u
2y dy=du
y dy= 1/2 du
Jadi,
∫▒y/((1+ y^2 )) dy=∫▒(1/2 du)/u=1/2 ∫▒du/u = 1/2 ln〖(u)+ c= 1/2〗 ln〖(1+ y^2 )+ c〗
Dengan demikian,
∫▒y/((1+ y^2 )) dy=∫▒2x/((1+2 x^2 )) dx
1/(2 ) ln〖(1+ y^2 )+ c_1 〗 = 1/2 ln(1+2x^2 )+ c_2
1/2 〖 ln〗〖(1+ y^2 ) - 〗 1/2 ln(1+2x^2 )=+ c_2- c_1
〖 ln〗〖(1+ y^2 ) - 〗 ln(1+2x^2 )=2C
〖 ln〗〖((1+ y^2 ))/((1+2x^2 ) ) = 〗 e^2c
x2y dx + (x +1) dy = 0
Jawab:
x^2 y dx+(x +1)dy= 0
(x +1)dy=-x^2 y dx Kedua ruas di kali 1/(y(x + 1))
1/y dy=-x^2/((x+1)) dx
∫▒1/y dy=-∫▒x^2/((x+1)) dx
Selesaikan terlebih dahulu,
∫▒x^2/((x+1)) dx=⋯
Misal:
x +1=u maka,x=u-1 dan x^2=〖(u-1)〗^2
dx=du
Jadi,
∫▒x^2/((x+1)) dx=∫▒〖(u-1)〗^2/u du
∫▒x^2/((x+1)) dx=∫▒((u^2-2u+1))/u du
∫▒x^2/((x+1)) dx=∫▒u^2/u du-∫▒2u/u du+∫▒1/u du
∫▒x^2/((x+1)) dx=∫▒u du-∫▒2 du+∫▒1/u du
∫▒x^2/((x+ 1)) dx= 1/2 u^2- 2u+ln〖(u)+c〗
∫▒x^2/((x+ 1)) dx= 1/2 〖(x+1)〗^2- 2(x + 1)+ ln〖(x + 1)+c〗
∫▒x^2/((x+ 1)) dx= 1/2 (x^(2 )+2x+1)- 2x-2+ ln〖(x + 1)+c〗
∫▒x^2/((x+ 1)) dx= 1/2 x^2+ x + 1/2 - 2x-2+ ln〖(x + 1)+c〗
∫▒x^2/((x+ 1)) dx= 1/2 x^2- x-3/2 + ln〖(x + 1)+c〗
Dengan demikian,
∫▒1/y dy=-∫▒x^2/((x+1)) dx
ln(y)+ c_1 =-( 1/2 x^2- x-3/2 + ln〖(x + 1) + c_2 )〗
ln(y)+ c_1 =- 1/2 x^2+ x+3/2 - ln〖(x + 1)- c_2 〗
ln〖(y)+〗 ln(x + 1) =- 1/2 x^2+ x+3/2 - c_2- c_1
ln(y)(x + 1)=- 1/2 x^2+ x+3/2 +C
y3 dx + √(1- x^2 ) dy = 0
Jawab:
y^3 dx+ √(1 - x^2 ) dy=0
√(1 - x^2 ) dy=〖-y〗^3 dx Kedua ruas di kali 1/(y^3 √(1 - x^2 ))
1/y^3 dy= -1/√(1- x^2 ) dx
∫▒1/y^3 dy= -∫▒1/√(1- x^2 ) dx
-1/2 y^(-2)+ c_1= -(arc sin〖x+ c_2)〗
-1/(2y^(2 ) )+ c_1= -arc sin〖x+ c_2 〗
1/(2y^(2 ) )+ arc sinx= c_2-c_1
1/y^(2 ) +2 arc sinx= 2C
1/y^(2 ) +2 arc sinx= C
dy/dx= 〖(2-y)〗^2/(2√(1+x))
Jawab:
dy/dx= 〖(2-y)〗^2/(2√(1+x))
2√(1+x) dy〖= (2-y)〗^2 dx Kedua ruas di kali dengan 1/(〖(2-y)〗^2 √(1+x))
2 1/〖(2-y)〗^2 dy= 1/√(1 +x) dx
∫▒〖2 1/〖(2-y)〗^2 〗 dy= ∫▒1/√(1 +x) dx
2∫▒1/〖(2-y)〗^2 dy= ∫▒1/√(1 +x) dx
Selesaikan satu persatu terlebih dahulu..
∫▒2/〖(2-y)〗^2 dy=⋯
Misal:
2-y=u
-dy=du
dy= -du
Jadi,
∫▒2/〖(2-y)〗^2 dy=2 ∫▒(-1)/u^2 du=2 ∫▒〖-u〗^(-2) du=2u^(-1)+ c= 21/u +c=2 1/((2-y))+ c
∫▒1/√(1 +x) dx=⋯
Misal:
1+x=u
dx=du
Jadi,
∫▒1/√(1 +x) dx= ∫▒1/u^(1/2) du= ∫▒〖u^(-1/2)= -2u^(1/2) 〗+ c= -2√(1+x) + c
Dengan demikian,
2∫▒1/〖(2-y)〗^2 dy= ∫▒1/√(1 +x) dx
2(1/((2-y) )+ c)= -2√(1+x) + c
2/(2-y)= -2√(1+x) + c
2/(2-y)+2√(1+x)=c_2 - c_1
2/(2-y)+2√(1+x)=C
dy/dx= (4x+x y^2)/(y-x^2 y)
Jawab:
dy/dx= (4x+x y^2)/(y-x^2 y)
(y-x^2 y)dy=(4x+〖xy〗^2 )dx
y(1-x^2 )dy=x(4+y^2 )dx Kedua ruas di kali 1/((1 - x^2 )(4+y^2))
y/((4+y^2 ) ) dy=x/((1 – x^2 ) ) dx
∫▒y/((4+y^2)) dy=∫▒x/((1 - x^2 ) ) dx
Selesaikan satu persatu..
∫▒y/((4+y^2)) dy=⋯
Misal:
4+y^2=u
2y dy=du
y dy= 1/2 du
Jadi,
∫▒y/((4+y^2)) dy=∫▒(1/2 du)/u=1/2 ∫▒du/u= 1/2 ln〖(u)+ c=〗 1/2 ln〖(4+y^2 )+ c〗
∫▒x/((1 – x^2 ) ) dx=⋯
Misal:
1-x^2=u
-2x dx=du
xdx= -1/2 du
Jadi,
∫▒x/((1 – x^2 ) ) dx= ∫▒(-1/2 du)/u= -1/2 ∫▒du/u=- 1/2 ln〖(u)+c〗= - 1/2 ln〖(1-x^2 )+c〗
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Dengan demikian,
∫▒y/((4+y^2)) dy=∫▒x/((1 - x^2 ) ) dx
1/2 ln〖(4+y^2 )+ c_1 〗= - 1/2 ln〖(1-x^2 )+c〗_2
1/2 ln〖(4+y^2 )+1/2 ln(1-x^2 )_ 〗= c_2- c_1
ln〖(4+y^2 )(1-x^2)〗= 2C
〖(4+y^2 )(1-x^2)〗= e^2c
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