Determine the nature and stability of the critical point (0,0) for the following system:

$\frac{dx}{dt}=x+2y+2\mathrm{sin}y$

$\frac{dy}{dt}=-3y-x{e}^{x}$

$\frac{dx}{dt}=x+2y+2\mathrm{sin}y$

$\frac{dy}{dt}=-3y-x{e}^{x}$

skilpadw3
2022-07-27
Answered

Determine the nature and stability of the critical point (0,0) for the following system:

$\frac{dx}{dt}=x+2y+2\mathrm{sin}y$

$\frac{dy}{dt}=-3y-x{e}^{x}$

$\frac{dx}{dt}=x+2y+2\mathrm{sin}y$

$\frac{dy}{dt}=-3y-x{e}^{x}$

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Jeroronryca

Answered 2022-07-28
Author has **13** answers

The equilibrium solutions (or points) to a system of first order differential equations are the points at which the first derivatives are equal to zero.

That is, for the system:

dx/dt = f(x,y)

dy/dt = g(x,y),

the equilibrium points are the solutions to the algebraic equations:

f(x,y) = 0

g(x,y) = 0

Now = 0

$\Rightarrow x+2y+2\mathrm{sin}y=0$

and =0

Therefore $-3y-x{e}^{x}=0$

From these two we observe that the critical point is (0,0)

That is, for the system:

dx/dt = f(x,y)

dy/dt = g(x,y),

the equilibrium points are the solutions to the algebraic equations:

f(x,y) = 0

g(x,y) = 0

Now = 0

$\Rightarrow x+2y+2\mathrm{sin}y=0$

and =0

Therefore $-3y-x{e}^{x}=0$

From these two we observe that the critical point is (0,0)

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