Answer:
Step-by-step explanation:
Given that,
y' = 17y ( 1-y^7)
Let y=1
Then, y' = 0 for all t
Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.
So, the graph solution will be
y(0) = 1 and this will be an horizontal line
If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.
Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.
So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.
This make the limit be 1.
This shows that x = 1 is a stable equilibrium.
Answer:
for part a i think it is the second one sorry if i do get it wrong and i couldnt answer the last one sorry
Step-by-step explanation:
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Answer:
Step-by-step explanation:
I'm going to assume you mean passes through the point (-2,5) since -2 and 5 could mean a lot of different things
To find the perpendicular line's gradient, you would know that m1*m2 = -1
In this case m1 = 7, so m2 must be -1/7
Now we substitute the gradient into the equation
y = -1/7x + b
To find b, we substitute (-2,5) into the equation above
5 = -1/7*-2+b
5 = 2/7 + b
b = 33/7
Therefore the equation of the line is
y = -1/7x+33/7
