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

$\frac{dx}{dt}=-\mathrm{sin}(x-y)$

$\frac{dy}{dt}=1-5y-{e}^{x}$

$\frac{dx}{dt}=-\mathrm{sin}(x-y)$

$\frac{dy}{dt}=1-5y-{e}^{x}$

Nathalie Fields
2022-07-24
Answered

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

$\frac{dx}{dt}=-\mathrm{sin}(x-y)$

$\frac{dy}{dt}=1-5y-{e}^{x}$

$\frac{dx}{dt}=-\mathrm{sin}(x-y)$

$\frac{dy}{dt}=1-5y-{e}^{x}$

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dominicsheq8

Answered 2022-07-25
Author has **15** answers

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

= 0

This implies that $x-y=0\Rightarrow x=y$

=0

Then the above Equation becomes

$1-5x-{e}^{x}=0$

$\Rightarrow {e}^{x}=1-5x$

when x = 0 both L.H.S and R.H.S are Equal

Therefore x = 0 and y = 0

Therefore 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

= 0

This implies that $x-y=0\Rightarrow x=y$

=0

Then the above Equation becomes

$1-5x-{e}^{x}=0$

$\Rightarrow {e}^{x}=1-5x$

when x = 0 both L.H.S and R.H.S are Equal

Therefore x = 0 and y = 0

Therefore the critical point is (0,0)

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