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

7. \(A\)as in exercise 5

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 5,

\(\begin{array}{l}A = \left( {\begin{array}{ {20}{c}}7&{ - 1}\\3&3\end{array}} \right)\\{v_1} = \left( {\begin{array}{ {20}{l}}1\\3\end{array}} \right){\rm{ and }}{{\rm{v}}_2} = \left( {\begin{array}{ {20}{l}}1\\1\end{array}} \right)\end{array}\)

With eigenvalues 4 and 6.

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

Also, consider\(D = \left( {\begin{array}{ {20}{l}}4&0\\0&6\end{array}} \right)\).

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

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

Here,\(P\)have 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 \(\left( {\begin{array}{ {20}{l}}{{y_1}^\prime (t)}\\{{y_1}^\prime (t)}\end{array}} \right) = \left( {\begin{array}{ {20}{l}}4&0\\0&6\end{array}} \right)\left( {\begin{array}{ {20}{l}}{{y_1}(t)}\\{{y_1}(t)}\end{array}} \right)\).