Answer:
y = tan (x+C) - x - 9
y = (x+C / 2)^2 + 3x - 2
y = e^(Cx -1) / x
Step-by-step explanation:
8 )
(dy/dx) = (x + y + 9)^2
Take substitution u = x + y + 9
du / dx = 1 + dy/dx
du / dx = 1 + u^2
Separating variables:
du / (1 + u^2) = dx
Integrating both sides
arctan(u) = x + C
u = tan (x+C)
Back substitution:
y = tan (x+C) - x - 9
Answer: y = tan (x+C) - x - 9
9)
(dy/dx) = 3 + sqrt(y - 3x + 2)
Take substitution u = y + 2 - 3x
du / dx = dy/dx - 3
du / dx = sqrt(u)
Separating variables:
du / sqrt(u) = dx
Integrating both sides
2*sqrt(u) = x + C
u = (x+C / 2)^2
Back substitution:
y = (x+C / 2)^2 + 3x - 2
Answer : y = (x+C / 2)^2 + 3x - 2
10)
x(dy/dx) = y ln(xy)
Take substitution u = xy
du / dx = x.dy/dx + y
du / dx = (u / x) * ( 1+ ln(u) )
Separating variables:
du / u*( 1+ln(u) ) = dx / x
Integrating both sides
Ln ( Ln (u) + 1 ) = Ln (x) + C
Ln (u) + 1 = C*x
u = e^(Cx -1)
Back substitution:
y = e^(Cx -1) / x
Answer: y = e^(Cx -1) / x