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

Question
Equations
Solve the given system of differential equations.
$$\displaystyle{Q}.{x}′_{{1}}={x}_{{1}}−{3}{x}_{{2}},{x}′_{{2}}={3}{x}_{{1}}+{x}_{{2}}$$.

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,
$$\displaystyle{x}_{{2}}{\left({t}\right)}={\left[\begin{array}{c} {e}^{{-{3}{i}+{1}}}\\{i}{e}^{{-{3}{i}+{1}}}\end{array}\right]}$$

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