How Do I Solve These Systems of Differential Equations with Specific Conditions?

[Tarhead]

I am having problems with solving systems of differential equations.

x'= [(-3 ) (gamma)]x
...[ ( 6 ) ( 4 ) ]

I am supposed tofind the interval of values of gamma for a) stable focus and b) stable node.

I started by
[(-3-r) (gamma)][x1] = [0]
[( 6 ) (4-r ) ][x2]...[0]

det(A-rI) = (-3-r)(4-r)-6(gamma) = 0
= r^2-r-12-6(gamma)= 0

but I don't know where to go after this point to find these different intervals.


For another problem:
x'= [0 3]x
...[-12 0] with initial conditions x1(0)= 1, x2(0) = 2

show that the solution x(t) is periodic and determine its period. Additionally to find the moment(s) when the point x(t) is closest to the equilibrium point 0.

For this I have
[(-r ) (3)][x1] = [0]
[(-12) ( -r)][x2]...[0]
so r^2 + 36 = 0

how do I or do I factor this? and after I find my values of r and plug them back in, where do I go?

 

[CrusaderSean]

for the first problem,
a) I'm going to assume stable focus means stable spiral... that occurs when you have complex roots and real part is negative
b) stable node occurs for real eigenvalues < 0 and det A (your system) is > 0
your book should have a section on stability criteria w/ critical parabola diagram. you just look at different regions of the graph to determine eigenvalues <-> phase portrait. qualitatively, i think its clear your real part of eigenvalue should be negative to obtain a stable solution. you can't have a exp(+rt) to be stable as t->infinity.

i'm learning diff eq myself and i haven't learned periodic solutions yet.. so can't help you w/ second problem.

 

[xanthym]

I am having problems with solving systems of differential equations.

x'= [(-3 ) (gamma)]x
...[ ( 6 ) ( 4 ) ]

I am supposed tofind the interval of values of gamma for a) stable focus and b) stable node.

I started by
[(-3-r) (gamma)][x1] = [0]
[( 6 ) (4-r ) ][x2]...[0]

det(A-rI) = (-3-r)(4-r)-6(gamma) = 0
= r^2-r-12-6(gamma)= 0

but I don't know where to go after this point to find these different intervals.


For another problem:
x'= [0 3]x
...[-12 0] with initial conditions x1(0)= 1, x2(0) = 2

show that the solution x(t) is periodic and determine its period. Additionally to find the moment(s) when the point x(t) is closest to the equilibrium point 0.

For this I have
[(-r ) (3)][x1] = [0]
[(-12) ( -r)][x2]...[0]
so r^2 + 36 = 0

how do I or do I factor this? and after I find my values of r and plug them back in, where do I go?

SOLUTION HINTS:
Problem #1:
Given following system of differential equations:

1: \ \ \ \ \ \left [<br /> \begin{array}{r}<br /> x_{1}^{&#039;} \\<br /> x_{2}^{&#039;} \\<br /> \end{array}<br /> \right ] \ \, = \ \, \left [<br /> \begin{array}{rr}<br /> -3 &amp; \gamma \\<br /> 6 &amp; 4 \\<br /> \end{array}<br /> \right ] \left [<br /> \begin{array}{r}<br /> x_{1}(t) \\<br /> x_{2}(t) \\<br /> \end{array}<br /> \right ]

Determine eigenvalues "λ":
(-3 - λ)*(4 - λ) - 6*γ = 0
::: ⇒ λ² - λ - 6(2 + γ) = 0
::: ⇒ λ = (+1/2) ± (1/2)*sqrt{1 + 24*(2 + γ)}

STABLE NODE occurs when BOTH "λ" solutions are real and NEGATIVE. There are NO VALUES of "γ" for which this is true for both "λ"s because of the "(+1/2)" first term above.
STABLE FOCUS occurs when both "λ" solutions are COMPLEX (α ± βi) with NEGATIVE real "α". There are NO VALUES of "γ" for which this is true because of the "(+1/2)" first term above.


Problem #2:
Given following system of differential equations:

2: \ \ \ \ \ \left [<br /> \begin{array}{r}<br /> x_{1}^{&#039;} \\<br /> x_{2}^{&#039;} \\<br /> \end{array}<br /> \right ] \ \, = \ \, \left [<br /> \begin{array}{rr}<br /> 0 &amp; 3 \\<br /> -12 &amp; 0 \\<br /> \end{array}<br /> \right ] \left [<br /> \begin{array}{r}<br /> x_{1}(t) \\<br /> x_{2}(t) \\<br /> \end{array}<br /> \right ]

Determine eigenvalues "λ":
λ² + 36 = 0
::: ⇒ λ = ±(6i)
::: ⇒ x(t) = exp{±(6i)*t}
::: ⇒ x(t) = cos(ω*t) ± i*sin(ω*t) where ω = 2*π*f = 6

Determine period "T" of sinusoidal solutions "x(t)" using relationship between frequency "f" and "T", which is {T = 1/f}.


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