(1 point) solve the separable differential equation y′(x)=2y(x)+18√, y′(x)=2y(x)+18, and find the particular solution satisfying the initial condition y(5)=9.

We write the equation in terms of dy/dx,
y'(x)=sqrt (2y(x)+18)
dy/dx = sqrt(2y + 18)
dy/dx = sqrt(2) ( sqrt(y + 9)) 

Separating the variables in the equation, we will have:
1/sqrt(y + 9) dy= sqrt(2) dx 

Integrating both sides, we will obtain
2sqrt(y+9) = x(sqrt(2)) + c 
where c is a constant and can be determined by using the boundary condition given 
y(5)=9 : x = 5, y = 9
sqrt(9+9) = 5/sqrt(2) + C 

C = sqrt(18) - 5/sqrt(2) = sqrt(2) / 2

Substituting to the original equation,
sqrt(y+9) = x/sqrt(2) + sqrt(2) / 2

sqrt(y+9) = (2x + 2) / 2sqrt(2) 

Squaring both sides, we will obtain,
y + 9 = ((2x+2)^2) / 8

y = ((2x+2)^2) / 8 - 9