Solve the given system of differential equations. Q.x′_1 =x_1−3x_2, x′_2 =3x_1+x_2.

Solve the given system of differential equations. Q.x′_1 =x_1−3x_2, x′_2 =3x_1+x_2.

Question
Equations
asked 2021-02-09
Solve the given system of differential equations.
\(\displaystyle{Q}.{x}′_{{1}}={x}_{{1}}−{3}{x}_{{2}},{x}′_{{2}}={3}{x}_{{1}}+{x}_{{2}}\).

Answers (1)

2021-02-10
Step 1
Provided a system of differential equations.
Arrange the system as shown below,
x'=Ax
\(\displaystyle{x}{\left({t}\right)}={\left[\begin{array}{c} {x}{1}{\left({t}\right)}\\{x}{2}{\left({t}\right)}\end{array}\right]}\)
\(\displaystyle{A}={\left[\begin{array}{cc} {1}&-{3}\\{3}&{1}\end{array}\right]}\)
Eigen values of A is given by \(\displaystyle{\left|{{A}-\lambda{I}}\right|}={0}\)
\(\displaystyle{\left({1}-\lambda\right)}^{{2}}+{9}={0}\)
\(\displaystyle{\left({1}-\lambda\right)}^{{2}}=-{9}\)
\(\displaystyle{\left({1}-\lambda\right)}=\pm{3}{i}\)
\(\displaystyle\lambda=\pm{3}{i}+{1}\)
Take \(\displaystyle\lambda={3}{i}+{1}\)
\(\displaystyle{A}-\lambda{I}{\left[\begin{array}{cc} {1}-{3}{i}-{1}&-{3}\\{3}&{1}-{3}{i}-{1}\end{array}\right]}\)
\(\displaystyle{A}-\lambda{I}={\left[\begin{array}{cc} -{3}{i}&-{3}\\{3}&-{3}{i}\end{array}\right]}\)
Step 2
Now
\(\displaystyle-{3}{i}{x}_{{1}}-{3}{x}_{{2}}={0}\)
\(\displaystyle-{3}{i}{x}_{{1}}={3}{x}_{{2}}\)
\(\displaystyle-{i}{x}_{{1}}={x}_{{2}}\)
Take Eigen vector \(\displaystyle{v}_{{1}}={\left[\begin{array}{c} {1}\\-{j}\end{array}\right]}\)
Then \(\displaystyle{x}{\left({t}\right)}={e}^{{\lambda{t}}}{v}_{{1}}={e}^{{{3}{i}+{1}}}{v}_{{1}}\)
\(\displaystyle{x}{\left({t}\right)}={e}^{{{3}{i}+{1}}}{\left[\begin{array}{c} {1}\\-{j}\end{array}\right]}\)
\(\displaystyle{x}_{{1}}{\left({t}\right)}={\left[\begin{array}{c} {e}^{{{3}{i}+{1}}}\\-{i}{e}^{{{3}{i}+{1}}}\end{array}\right]}\)
As both the Eigen values are conjugate hence,
\(\displaystyle{x}_{{2}}{\left({t}\right)}={\left[\begin{array}{c} {e}^{{-{3}{i}+{1}}}\\{i}{e}^{{-{3}{i}+{1}}}\end{array}\right]}\)
0

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