A.
1.
x β X + h , y β Y + k x\to X+h, y\to Y+k x β X + h , y β Y + k
2 h β 3 k β 4 = 0 2h-3k-4=0 2 h β 3 k β 4 = 0
3 h β 4 k β 2 = 0 3h-4k-2=0 3 h β 4 k β 2 = 0
h = k β 2 h=k-2 h = k β 2
2 k β 4 β 3 k β 4 = 0 2k-4-3k-4=0 2 k β 4 β 3 k β 4 = 0
k = β 8 , h = β 10 k=-8,h=-10 k = β 8 , h = β 10
x = X β 10 , y = Y β 8 x=X-10,y=Y-8 x = X β 10 , y = Y β 8
d Y d X = 2 X β 3 Y 3 X β 4 Y \frac{dY}{dX}=\frac{2X-3Y}{3X-4Y} d X d Y β = 3 X β 4 Y 2 X β 3 Y β
Y = t X , Y β² = t β² X + t Y=tX,Y'=t'X+t Y = tX , Y β² = t β² X + t
t β² X + t = 2 X β 3 t X 3 X β 4 t X t'X+t=\frac{2X-3tX}{3X-4tX} t β² X + t = 3 X β 4 tX 2 X β 3 tX β
t β² X = 2 β 3 t β 3 t + 4 t 2 3 β 4 t t'X=\frac{2-3t-3t+4t^2}{3-4t} t β² X = 3 β 4 t 2 β 3 t β 3 t + 4 t 2 β
3 β 4 t 4 t 2 β 6 t + 2 d t = d X X \frac{3-4t}{4t^2-6t+2}dt=\frac{dX}{X} 4 t 2 β 6 t + 2 3 β 4 t β d t = X d X β
β l n ( 2 t 2 β 3 t + 1 ) 2 = l n X + l n c 1 -\frac{ln(2t^2-3t+1)}{2}=lnX+lnc_1 β 2 l n ( 2 t 2 β 3 t + 1 ) β = l n X + l n c 1 β
1 2 t 2 β 3 t + 1 = c 2 X 2 \frac{1}{2t^2-3t+1}=c_2X^2 2 t 2 β 3 t + 1 1 β = c 2 β X 2
1 2 ( Y / X ) 2 β 3 ( Y / X ) + 1 = c 2 X 2 \frac{1}{2(Y/X)^2-3(Y/X)+1}=c_2X^2 2 ( Y / X ) 2 β 3 ( Y / X ) + 1 1 β = c 2 β X 2
2 ( y + 8 x + 10 ) 2 β 3 y + 8 x + 10 + 1 = c ( x + 10 ) 2 2(\frac{y+8}{x+10})^2-3\frac{y+8}{x+10}+1=\frac{c}{(x+10)^2} 2 ( x + 10 y + 8 β ) 2 β 3 x + 10 y + 8 β + 1 = ( x + 10 ) 2 c β
2.
x β X + h , y β Y + k x\to X+h, y\to Y+k x β X + h , y β Y + k
2 h β k β 3 = 0 2h-k-3=0 2 h β k β 3 = 0
h + 4 k + 3 = 0 h+4k+3=0 h + 4 k + 3 = 0
h = 5 k + 6 h=5k+6 h = 5 k + 6
9 k + 9 = 0 9k+9=0 9 k + 9 = 0
k = β 1 , h = 1 k=-1,h=1 k = β 1 , h = 1
x = X + 1 , y = Y β 1 x=X+1,y=Y-1 x = X + 1 , y = Y β 1
d Y d X = 2 X β Y X + 4 Y \frac{dY}{dX}=\frac{2X-Y}{X+4Y} d X d Y β = X + 4 Y 2 X β Y β
Y = t X , Y β² = t β² X + t Y=tX,Y'=t'X+t Y = tX , Y β² = t β² X + t
t β² X + t = 2 X β t X X + 4 t X = 2 β t 1 + 4 t t'X+t=\frac{2X-tX}{X+4tX}=\frac{2-t}{1+4t} t β² X + t = X + 4 tX 2 X β tX β = 1 + 4 t 2 β t β
t β² X = β 4 t 2 β 2 t + 2 1 + 4 t t'X=\frac{-4t^2-2t+2}{1+4t} t β² X = 1 + 4 t β 4 t 2 β 2 t + 2 β
1 + 4 t β 4 t 2 β 2 t + 2 d t = d X X \frac{1+4t}{-4t^2-2t+2}dt=\frac{dX}{X} β 4 t 2 β 2 t + 2 1 + 4 t β d t = X d X β
β l n ( 2 t 2 + t β 1 ) 2 = l n X + l n c 1 -\frac{ln(2t^2+t-1)}{2}=lnX+lnc_1 β 2 l n ( 2 t 2 + t β 1 ) β = l n X + l n c 1 β
1 2 t 2 + t β 1 = c 2 X 2 \frac{1}{2t^2+t-1}=c_2X^2 2 t 2 + t β 1 1 β = c 2 β X 2
1 2 ( Y / X ) 2 + ( Y / X ) β 1 = c 2 X 2 \frac{1}{2(Y/X)^2+(Y/X)-1}=c_2X^2 2 ( Y / X ) 2 + ( Y / X ) β 1 1 β = c 2 β X 2
2 ( y + 1 x β 1 ) 2 + y + 1 x β 1 β 1 = c ( x β 1 ) 2 2(\frac{y+1}{x-1})^2+\frac{y+1}{x-1}-1=\frac{c}{(x-1)^2} 2 ( x β 1 y + 1 β ) 2 + x β 1 y + 1 β β 1 = ( x β 1 ) 2 c β
3.
x β y = v x-y=v x β y = v
1 β d v d x = v + 2 v β 6 1-\frac{dv}{dx}=\frac{v+2}{v-6} 1 β d x d v β = v β 6 v + 2 β
β 8 v β 6 = d v d x -\frac{8}{v-6}=\frac{dv}{dx} β v β 6 8 β = d x d v β
β 8 x = ( v β 6 ) 2 2 + c -8x=\frac{(v-6)^2}{2}+c β 8 x = 2 ( v β 6 ) 2 β + c
β 8 x = ( x β y β 6 ) 2 2 + c -8x=\frac{(x-y-6)^2}{2}+c β 8 x = 2 ( x β y β 6 ) 2 β + c
4.
x β 2 y = v x-2y=v x β 2 y = v
1 2 ( 1 β d v d x ) = β v + 4 2 v β 1 \frac{1}{2}(1-\frac{dv}{dx})=-\frac{v+4}{2v-1} 2 1 β ( 1 β d x d v β ) = β 2 v β 1 v + 4 β
1 + 2 v + 8 v β 6 = d v d x 1+\frac{2v+8}{v-6}=\frac{dv}{dx} 1 + v β 6 2 v + 8 β = d x d v β
3 v + 2 v β 6 = d v d x \frac{3v+2}{v-6}=\frac{dv}{dx} v β 6 3 v + 2 β = d x d v β
d x = v β 6 3 v + 2 d v dx=\frac{v-6}{3v+2}dv d x = 3 v + 2 v β 6 β d v
x = 3 v β 20 l n ( 3 v + 2 ) 9 + c x=\frac{3v-20ln(3v+2)}{9}+c x = 9 3 v β 20 l n ( 3 v + 2 ) β + c
x = 3 ( x β 2 y ) β 20 l n ( 3 ( x β 2 y ) + 2 ) 9 + c x=\frac{3(x-2y)-20ln(3(x-2y)+2)}{9}+c x = 9 3 ( x β 2 y ) β 20 l n ( 3 ( x β 2 y ) + 2 ) β + c
5.
x β X + h , y β Y + k x\to X+h, y\to Y+k x β X + h , y β Y + k
2 h β k β 3 = 0 2h-k-3=0 2 h β k β 3 = 0
h + 4 k + 3 = 0 h+4k+3=0 h + 4 k + 3 = 0
h = 5 k + 6 h=5k+6 h = 5 k + 6
9 k + 9 = 0 9k+9=0 9 k + 9 = 0
k = β 1 , h = 1 k=-1,h=1 k = β 1 , h = 1
x = X + 1 , y = Y β 1 x=X+1,y=Y-1 x = X + 1 , y = Y β 1
d Y d X = X + 4 Y 2 X β Y \frac{dY}{dX}=\frac{X+4Y}{2X-Y} d X d Y β = 2 X β Y X + 4 Y β
Y = t X , Y β² = t β² X + t Y=tX,Y'=t'X+t Y = tX , Y β² = t β² X + t
t β² X + t = X + 4 t X 2 X β t X = 1 + 4 t 2 β t t'X+t=\frac{X+4tX}{2X-tX}=\frac{1+4t}{2-t} t β² X + t = 2 X β tX X + 4 tX β = 2 β t 1 + 4 t β
t β² X = β t 2 + 2 t + 1 1 + 4 t t'X=\frac{-t^2+2t+1}{1+4t} t β² X = 1 + 4 t β t 2 + 2 t + 1 β
1 + 4 t β t 2 + 2 t + 1 d t = d X X \frac{1+4t}{-t^2+2t+1}dt=\frac{dX}{X} β t 2 + 2 t + 1 1 + 4 t β d t = X d X β
5 2 2 l n ( t + 2 β 1 t β 2 β 1 ) β 2 l n ( t 2 β 2 t β 1 ) = l n X + l n c \frac{5}{2\sqrt 2}ln(\frac{t+\sqrt 2-1}{t-\sqrt 2-1})-2ln(t^2-2t-1)=lnX+lnc 2 2 β 5 β l n ( t β 2 β β 1 t + 2 β β 1 β ) β 2 l n ( t 2 β 2 t β 1 ) = l n X + l n c
5 2 2 l n ( Y / X + 2 β 1 Y / X β 2 β 1 ) β 2 l n ( ( Y / X ) 2 β 2 Y / X β 1 ) = l n ( c X ) \frac{5}{2\sqrt 2}ln(\frac{Y/X+\sqrt 2-1}{Y/X-\sqrt 2-1})-2ln((Y/X)^2-2Y/X-1)=ln(cX) 2 2 β 5 β l n ( Y / X β 2 β β 1 Y / X + 2 β β 1 β ) β 2 l n (( Y / X ) 2 β 2 Y / X β 1 ) = l n ( c X )
5 2 2 l n ( ( y + 1 ) / ( x β 1 ) + 2 β 1 ( y + 1 ) / ( x β 1 ) β 2 β 1 ) β 2 l n ( ( ( y + 1 ) / ( x β 1 ) ) 2 β 2 ( y + 1 ) / ( x β 1 ) β 1 ) = \frac{5}{2\sqrt 2}ln(\frac{(y+1)/(x-1)+\sqrt 2-1}{(y+1)/(x-1)-\sqrt 2-1})-2ln(((y+1)/(x-1))^2-2(y+1)/(x-1)-1)= 2 2 β 5 β l n ( ( y + 1 ) / ( x β 1 ) β 2 β β 1 ( y + 1 ) / ( x β 1 ) + 2 β β 1 β ) β 2 l n ((( y + 1 ) / ( x β 1 ) ) 2 β 2 ( y + 1 ) / ( x β 1 ) β 1 ) =
= l n ( c ( x β 1 ) ) =ln(c(x-1)) = l n ( c ( x β 1 ))
B.
1.
( π€ 3 + π€ π§ 2 β π§ ) z = 2 z w β 1 (π€^3 + π€π§ ^2 β π§)_z=2zw-1 ( w 3 + w z 2 β z ) z β = 2 z w β 1
( π§ 2 + π€ 2 π§ β π€ ) w = 2 z w β 1 (π§ ^2 + π€^2 π§ β π€)_w=2zw-1 ( z 2 + w 2 z β w ) w β = 2 z w β 1
F = β« ( π€ 3 + π€ π§ 2 β π§ ) d w = w 4 / 4 + w 2 z 2 / 2 β z w + g ( z ) F=\int (π€^3 + π€π§ ^2 β π§)dw=w^4/4+w^2z^2/2-zw+g(z) F = β« ( w 3 + w z 2 β z ) d w = w 4 /4 + w 2 z 2 /2 β z w + g ( z )
F z = w 2 z β w + g β² ( z ) = π§ 2 + π€ 2 π§ β π€ F_z=w^2z-w+g'(z)=π§ ^2 + π€^2 π§ β π€ F z β = w 2 z β w + g β² ( z ) = z 2 + w 2 z β w
g β² ( z ) = π§ 2 g'(z)=π§ ^2 g β² ( z ) = z 2
g ( z ) = β« π§ 2 d z = z 3 / 3 + c g(z)=\intπ§ ^2dz=z^3/3+c g ( z ) = β« z 2 d z = z 3 /3 + c
F = w 4 / 4 + w 2 z 2 / 2 β z w + z 3 / 3 + c F=w^4/4+w^2z^2/2-zw+z^3/3+c F = w 4 /4 + w 2 z 2 /2 β z w + z 3 /3 + c
w 4 / 4 + w 2 z 2 / 2 β z w + z 3 / 3 + c = 0 w^4/4+w^2z^2/2-zw+z^3/3+c=0 w 4 /4 + w 2 z 2 /2 β z w + z 3 /3 + c = 0
2.
( c o s 2 π¦ β 3 π₯ 2 π¦ 2 ) y = β 2 s i n 2 y β 6 x 2 y (cos 2π¦ β 3π₯^ 2π¦ ^2 )_y=-2sin2y-6x^2y ( cos 2 y β 3 x 2 y 2 ) y β = β 2 s in 2 y β 6 x 2 y
( c o s 2 π¦ β 2 π₯ s i n 2 π¦ β 2 π₯ 3 π¦ ) x = β 2 s i n 2 y β 6 x 2 y (cos 2π¦ β 2π₯ sin 2π¦ β 2π₯ ^3π¦)_x=-2sin2y-6x^2y ( cos 2 y β 2 x s in 2 y β 2 x 3 y ) x β = β 2 s in 2 y β 6 x 2 y
F = β« ( c o s 2 π¦ β 3 π₯ 2 π¦ 2 ) d x = x c o s 2 y β x 3 y 2 + g ( y ) F=\int (cos 2π¦ β 3π₯^ 2π¦ ^2 )dx=xcos2y-x^3y^2+g(y) F = β« ( cos 2 y β 3 x 2 y 2 ) d x = x cos 2 y β x 3 y 2 + g ( y )
F y = β 2 x s i n 2 y β 2 y x 3 + g β² ( y ) = c o s 2 π¦ β 2 π₯ s i n 2 π¦ β 2 π₯ 3 π¦ F_y=-2xsin2y-2yx^3+g'(y)=cos 2π¦ β 2π₯ sin 2π¦ β 2π₯ ^3π¦ F y β = β 2 x s in 2 y β 2 y x 3 + g β² ( y ) = cos 2 y β 2 x s in 2 y β 2 x 3 y
g β² ( y ) = c o s 2 π¦ g'(y)=cos 2π¦ g β² ( y ) = cos 2 y
g ( y ) = β« c o s 2 y d y = s i n 2 y / 2 + c g(y)=\intop cos2ydy=sin2y/2+c g ( y ) = β« cos 2 y d y = s in 2 y /2 + c
F = x c o s 2 y β x 3 y 2 + s i n 2 y / 2 + c F=xcos2y-x^3y^2+sin2y/2+c F = x cos 2 y β x 3 y 2 + s in 2 y /2 + c
x c o s 2 y β x 3 y 2 + s i n 2 y / 2 + c = 0 xcos2y-x^3y^2+sin2y/2+c=0 x cos 2 y β x 3 y 2 + s in 2 y /2 + c = 0
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