Find the equilibrium points of the following nonlinear...

Find the equilibrium points of the following nonlinear system. Find its linearization at each equilibrium point. Determine the types of equilibrium points and investigate the stability. x' = y y' = -6x - y - 3x^2. Find the approximate solution of the system for the initial conditions x(0) = -2.1, y(0) = 0.1

Transcript

00:01 Hello everyone so let's look into this question find the equilibrium point of the following non linear system find the linearization at each equilibrium point determine the type of equilibrium points and investigate the stability find the approximate solution of the system for the initial condition x is not is equal to 2 .1 and y0 is equal to 0 .1 okay so x prime is equal to zy and y prime is going to minus six x minus y minus three x is y at equilibrium point so at equilibrium point we are going to see x not x prime is equal to zero and y prime is equal to zero so therefore therefore y is equal to zero and minus six x minus y minus three x square is equal to zero since since we know y is equal to zero so this is going to be minus six x minus six x minus three x squared is equal to 0 or we can say 3x square 1 second x square x square plus 2x is equal to 0 so x x plus 2 is equal to 0 x is equal to minus 2 so we have point 0 is 0 and minus 2 and 0 so these are the points we have to find out so f we have 2 function that is f is equal to y and g is equal to minus six x minus y minus three x square so d f by d x is equal to zero d f by d y is equal to one so and d g by d x is equal to minus six minus six x and d g by d y is going to be minus one so the matrix is going to be 0 1 minus 6 minus 6 x and minus 1 okay so for 0 0 0 0 0 0 minus x minus 1 and x and y so this is going to be 01 minus x minus 1 and y so this is going to be so this is the linear linear realization of 4 x prime is equal to y and y 5 and is equal to minus 6x minus y okay for minus 2 gamma 0 so we are going to get substitute this value to the equation we are going to get again so minus 6 6 is going to be minus 6 plus 12 this is going to be 6 0 1 6 minus 1 x and y so we are going to x tass is equal to y and y ds is equal to 6x minus y so this is going to be the linearization for the other one so in this question we had to find out determined the type of equilibrium points and investigate the stability all right so we have been so equilibrium point we have already found out so now we have to tag and stability so this is going to be for zero zero a is equal to 0 1 minus 6 minus 1 or we can say minus lambda 1 minus 6 minus lambda is equal to 0 so here we are going to get lambda is 4 plus lambda plus 6 is equal to 0 so lambda 1 and 2 is equal to minus i 23 divided 2 so 0 comma 0 is stable.

04:37 So minus 2 comma 0.

04:39 So a is going to be 061 minus 1.

04:44 Again, here we are going to characteristic equation 6...

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