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

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In summary, ~~The author is having difficulty solving systems of differential equations. They started by finding the eigenvalues, but are lost after that. For another problem, they are trying to find a periodic solution with a period. They found that the solution is periodic and determined its period. Additionally, they are trying to find the moments when the point x(t) is closest to the equilibrium point 0. They found that r^2 + 36 = 0 and how to factor it. However, for the first problem, they are unable to find a stable focus or a stable node. The book suggested looking for regions of the graph where eigenvalues are less than or equal to a certain value. They found that both solutions had
  • #1
Tarhead
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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?
 
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  • #2
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.
 
  • #3
Tarhead said:
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:

[tex] 1: \ \ \ \ \ \left [
\begin{array}{r}
x_{1}^{'} \\
x_{2}^{'} \\
\end{array}
\right ] \ \, = \ \, \left [
\begin{array}{rr}
-3 & \gamma \\
6 & 4 \\
\end{array}
\right ] \left [
\begin{array}{r}
x_{1}(t) \\
x_{2}(t) \\
\end{array}
\right ] [/tex]

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:

[tex] 2: \ \ \ \ \ \left [
\begin{array}{r}
x_{1}^{'} \\
x_{2}^{'} \\
\end{array}
\right ] \ \, = \ \, \left [
\begin{array}{rr}
0 & 3 \\
-12 & 0 \\
\end{array}
\right ] \left [
\begin{array}{r}
x_{1}(t) \\
x_{2}(t) \\
\end{array}
\right ] [/tex]

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}.


~~
 
Last edited:

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

1. What is a system of differential equations?

A system of differential equations is a set of equations that describe the relationship between multiple variables and their rates of change over time. It is used to model dynamic systems in various fields of science and engineering.

2. How is a system of differential equations solved?

A system of differential equations is typically solved using analytical or numerical methods. Analytical solutions involve finding an exact formula for the solution, while numerical methods use algorithms to approximate the solution.

3. What are the applications of a system of differential equations?

A system of differential equations has various applications in science and engineering, including in physics, chemistry, biology, economics, and engineering. It is commonly used to model and predict the behavior of complex systems.

4. How do you determine the stability of a system of differential equations?

The stability of a system of differential equations is determined by analyzing the behavior of its solutions over time. A system is considered stable if its solutions approach a steady state or oscillate around a fixed point, and unstable if its solutions become unbounded or diverge.

5. Can a system of differential equations have multiple solutions?

Yes, a system of differential equations can have multiple solutions. In some cases, a system may have an infinite number of solutions, and in other cases, it may have a unique solution. The number and nature of solutions depend on the initial conditions and parameters of the system.

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