The problem question is: dydt=ay+by2 we have to sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable.Thing is, how do you get the critical points?

Question

The problem question is:  dydt=ay+by2  we have to sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable.Thing is, how do you get the critical points?

okomgcae · Accepted Answer

dydt=f(y)=ay+by2 we take a=2 and b=4 and draw the graph of f(y) versus y  for the critical points we put f(y)=0 ay+by2=0 y(a+by)=0 y=0 and y=−ab (critical points) Now we find second derivative that is d2ydt2=adydt+2bydydt=(a+2by)(dydt)=(a+2by)(ay+by2) Second derivative will be zero when y=−a2b,y=0,y=−ab But y=0 and y=−ab are the solution of the equation so wetake y=−a2b ( inflection point) Now from the graph drawn above, we have dydt&amp;gt;0 for y&amp;lt;−ab , so y is increasing ddt&amp;lt;0 for −ab dydt&amp;gt;0 for y&amp;gt;0 so y is increasing and concavity changes at y=−a2b The graphs of solutions of the equation must have the genral shape as follows  Since the solutions diverge from the line y=0 and converge to the line y=−ab so y=0 is an unstable point and y=−ab is stable

The problem question is: \frac{dy}{dt}=ay+by^2 we have to sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable.Thing is, how do you get the critical points?

Answered question

Answer & Explanation

New Questions in Calculus 2