Write down all the equilibrium solutions of dy/dx = 2y^2 +3y + 1

Question

Show all working out please

Answer 1

Just let RHS be 0.

 

2y² + 3y + 1 = 0

 

(2y + 1)(y + 1) = 0

 

y = -1/2, y = -1

 

We can also see these from the general solution to the ODE.

 

dy/dx = 2y² + 3y + 1

 

dy/(2y² + 3y + 1) = dx

 

2dy/(2y + 1) - dy/(y + 1) = dx

 

ln|2y+1| - ln|y + 1| = x + C

 

ln|2 - 1/(y+1)| = x + C

 

2 - 1/(y + 1) = Ae^x

 

y = 1/(2 - Ae^x) - 1

 

lim_x→-∞ [1/(2 - Ae^x) - 1] = -1/2

 

lim_x→∞ [1/(2 - Ae^x) - 1] = -1