Finding specific solutions of a system of differential equations without computations
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I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.
I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.
I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.
What am I missing here?
differential-equations
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I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.
I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.
I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.
What am I missing here?
differential-equations
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.
I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.
I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.
What am I missing here?
differential-equations
I have a reflection matrix $A=begin{pmatrix}-frac{1}{2} & frac{sqrt{3}}{2} \ frac{sqrt{3}}{2} & frac{1}{2} end{pmatrix}$ and a linear, autonomous, and homogeneous system $frac{doverrightarrow{w}}{dt} = Aoverrightarrow{w}$, $overrightarrow{w}(t) = begin{pmatrix} x(t) \ y(t) end{pmatrix}$.
I've noticed that $left{begin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2}end{pmatrix},begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2}end{pmatrix}right}$ is an eigenbasis with eigenvalues $lambda = 1$ and $lambda = -1$. Thus, I thought that a general solution would be somewhere along the lines of $begin{pmatrix} x'(t) \ y'(t) end{pmatrix} = c_1e^tbegin{pmatrix}frac{1}{2} \ frac{sqrt{3}}{2} end{pmatrix} + c_2e^{-t}begin{pmatrix} -frac{sqrt{3}}{2} \ frac{1}{2} end{pmatrix}$.
I am unsure about how to go from here to finding specific solutions. I'm asked to find two solutions to this system "without performing calculations", and to indicate where the solutions curves start in the plane. Furthermore, I'm asked to find a solution that starts at $begin{pmatrix} 0 \ 5 end{pmatrix}$.
What am I missing here?
differential-equations
differential-equations
edited 20 hours ago
asked 21 hours ago
Dominic Hicks
826
826
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1 Answer
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$$A^{-1}=A,$$
$$e^{At}=A.
begin{pmatrix}{e^t} & 0\
0 & {e^{-t}}end{pmatrix}
.A^{-1}$$
Final solution is
$$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
=begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$A^{-1}=A,$$
$$e^{At}=A.
begin{pmatrix}{e^t} & 0\
0 & {e^{-t}}end{pmatrix}
.A^{-1}$$
Final solution is
$$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
=begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
add a comment |
up vote
1
down vote
$$A^{-1}=A,$$
$$e^{At}=A.
begin{pmatrix}{e^t} & 0\
0 & {e^{-t}}end{pmatrix}
.A^{-1}$$
Final solution is
$$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
=begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
$$A^{-1}=A,$$
$$e^{At}=A.
begin{pmatrix}{e^t} & 0\
0 & {e^{-t}}end{pmatrix}
.A^{-1}$$
Final solution is
$$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
=begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$
$$A^{-1}=A,$$
$$e^{At}=A.
begin{pmatrix}{e^t} & 0\
0 & {e^{-t}}end{pmatrix}
.A^{-1}$$
Final solution is
$$overrightarrow{w}(t) =e^{At}.begin{pmatrix} 0 \ 5 end{pmatrix}
=begin{pmatrix}frac{5 sqrt{3}, {{e}^{t}}}{4}-frac{5 sqrt{3}, {{e}^{-t}}}{4}\
frac{15 {{e}^{t}}}{4}+frac{5 {{e}^{-t}}}{4}end{pmatrix} $$
answered 19 hours ago
Aleksas Domarkas
6655
6655
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
add a comment |
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
1
1
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
Or alternatively, because of $A^2=I$, sorting the power series for even and odd powers gives $e^{tA}=cosh(t)I+sinh(t)A$.
– LutzL
18 hours ago
add a comment |
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