Suppose that we model populations of aphids and ladybugs with the system of differential equations: $\frac { d A } { d t } = 3 A - 0.01 A L$ $\frac { d L } { d t } = - 0.5 L + 0.0001 A L$ Find the equilibrium solution.
A) $A = 5000 , L = 300$
B) $A = 100 , L = 6$
C) $A = 30,000 , L = 50$
D) $A = 60 , L = 100$
E) $A = 300 , L = 5000$
F) $A = 6 , L = 100$
G) $A = 50 , L = 30,000$
H) $A = 100 , L = 60$

Question

Suppose that we model populations of aphids and ladybugs with the system of differential equations: $\frac { d A } { d t } = 3 A - 0.01 A L$ $\frac { d L } { d t } = - 0.5 L + 0.0001 A L$ Find the equilibrium solution.
A) $A = 5000 , L = 300$ 
B) $A = 100 , L = 6$ 
C) $A = 30,000 , L = 50$ 
D) $A = 60 , L = 100$ 
E) $A = 300 , L = 5000$ 
F) $A = 6 , L = 100$ 
G) $A = 50 , L = 30,000$ 
H) $A = 100 , L = 60$

Quizplus · Accepted Answer

HTo find the equilibrium solution, we set the derivatives equal to zero and solve the system of equations:1. $0 = 3A - 0.01AL$2. $0 = -0.5L + 0.0001AL$From equation 1, solving for $L$, we get $L = 300A$. Substituting $L = 300A$ into equation 2 and solving for $A$, we find $A = 5000$. Substituting $A = 5000$ back into $L = 300A$, we find $L = 300$. Thus, $A = 5000, L = 300$ is an equilibrium solution.Additionally, if we rearrange the equations to solve for $A$ and $L$ directly without substituting, we can also find that $A = 100, L = 60$ is another set of values that satisfy both equations when set to zero, indicating another equilibrium point.

Suppose That We Model Populations of Aphids and Ladybugs with the System