In Exercises 7 and 8, make a change of variable that decouples the equation\(x' = Ax\). Write the equation\(x\left( t \right) = Py\left( t \right)\)and show the calculation that leads to the uncoupled system\(y' = Dy\), specifying\(P\)and\(D\).

The general solutionfor any system of differential equations withthe eigenvalues\({\lambda _1}\)and\({\lambda _2}\)with the respective eigenvectors\({v_1}\)and\({v_2}\)is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here \({c_1}\) and \({c_2}\) are the constants from the initial condition.

From Exercise 6,

\(\begin{aligned}{l}A = \left( {\begin{aligned}{ {20}{l}}1&{ - 2}\\3&{ - 4}\end{aligned}} \right)\\{v_1} = \left( {\begin{aligned}{ {20}{l}}2\\3\end{aligned}} \right){\rm{ and }}{{\rm{v}}_2} = \left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right)\end{aligned}\)

With eigenvalues\( - 2\), and\( - 1\).

In order to decouple the equation\(x' = Ax\), such that,\(A = PD{P^{ - 1}}\)and\(D = {P^{ - 1}}AP\), consider\(P = \left( {\begin{aligned}{ {20}{l}}{{v_1}}&{{v_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}2&1\\3&1\end{aligned}} \right)\).

Also, let\(D = \left( {\begin{aligned}{ {20}{l}}{ - 2}&0\\0&{ - 1}\end{aligned}} \right)\).

Substituting\(x(t) = Py(t)\)into\(x' = Ax\), we have:

\(\begin{aligned}{c}\frac{d}{{dt}}(Py) = A(Py)\\ = PD{P^{ - 1}}(Py)\\ = PDy\end{aligned}\)

Here,\(P\)is having constant entries, so:

\(\frac{d}{{dt}}(Py) = P\frac{d}{{dt}}(y)\)

From the above two solutions, we have:

\(P\frac{d}{{dt}}(y) = PDy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}{P^{ - 1}}{\rm{ yields}} \to y' = Dy} \right\}\)

Hence, the solution is \(\left( {\begin{aligned}{ {20}{l}}{y_1^\prime (t)}\\{y_2^\prime (t)}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{r}}{ - 2}&0\\0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{y_1}(t)}\\{{y_2}(t)}\end{aligned}} \right)\).