(1/y)-(1/5y)+(1/(y+1))=(5/2y)

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Solution for (1/y)-(1/5y)+(1/(y+1))=(5/2y) equation:


D( y )

y = 0

y+1 = 0

y = 0

y = 0

y+1 = 0

y+1 = 0

y+1 = 0 // - 1

y = -1

y in (-oo:-1) U (-1:0) U (0:+oo)

t_1 = 0

1/(y+1)+t_1-1/5*y-5/2*y+y^-1 = 0

1/(y+1)+t_1-y/5+(-5*y)/2+1/y = 0

(1*2*5*y)/(2*5*y*(y+1))+(2*5*t_1*y*(y+1))/(2*5*y*(y+1))+(2*(-y)*y*(y+1))/(2*5*y*(y+1))+(-5*5*y*y*(y+1))/(2*5*y*(y+1))+(1*2*5*(y+1))/(2*5*y*(y+1)) = 0

2*5*t_1*y*(y+1)+2*(-y)*y*(y+1)-5*5*y*y*(y+1)+1*2*5*(y+1)+1*2*5*y = 0

10*t_1*y^2+10*t_1*y-2*y^3-25*y^3-2*y^2-25*y^2+10*y+10*y+10 = 0

10*t_1*y^2+10*t_1*y-27*y^3-27*y^2+10*y+10*y+10 = 0

10*t_1*y^2+10*t_1*y-27*y^3-27*y^2+20*y+10 = 0

10*t_1*y^2+10*t_1*y-27*y^3-27*y^2+20*y+10 = 0

10*t_1*y^2+10*t_1*y-27*y^3-27*y^2+20*y+10

10*t_1*y*(y+1)-27*y^3-27*y^2+20*y+10

10*t_1*y*(y+1)-27*y^2*(y+1)+20*y+10

10*(2*y+1)-27*y^2*(y+1)+10*t_1*y*(y+1)

(10*t_1*y-27*y^2)*(y+1)+10*(2*y+1)

11*(10*t_1*y-27*y^2)

(11*(10*t_1*y-27*y^2))/(2*5*y*(y+1)) = 0

(11*(10*t_1*y-27*y^2))/(2*5*y*(y+1)) = 0 // * 2*5*y*(y+1)

11*(10*t_1*y-27*y^2) = 0

10*t_1*y-27*y^2 = 0

y*(10*t_1-27*y) = 0

10*t_1-27*y = 0 // - 10*t_1

-27*y = -(10*t_1) // : -27

y = (-(10*t_1))/(-27)

y = (-10*t_1)/(-27)

y*(y-10/27*t_1) = 0

11*y*(y-10/27*t_1) = 0

( y-10/27*t_1 )

y-10/27*t_1 = 0 // + -10/27*t_1

y = -(-10/27*t_1)

y = 10/27*t_1

( y )

y = 0

y in { 0*10/27}

y in { 0}

y belongs to the empty set

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