Solve the given differential equations by using an appropriate substitution. the de is of the form (dy/dx) = f(ax + by + c), which is given in (5) of section 2.5.
8) (dy/dx) = (x + y + 9)2
9) (dy/dx) = 3 + sqrt(y - 3x + 2)

determine an appropriate substitution and then solve the differential equation.
10) xy' = y ln(xy)

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

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

y = e^(Cx -1) / x

whats the question?

step-by-step explanation: