Question

In Exercises 7 and 8, make a change of variable that decouples the equation$x^{\prime }=Ax$. Write the equation$x\left(t\right)=Py\left(t\right)$and show the calculation that leads to the uncoupled system$y^{\prime }=Dy$, specifying$P$and$D$.

Short answer

The required equation is:

$\left(\begin{array}{r}20l{y}_1^{\prime }(t)\\ y_2^{\prime }(t)\end{array}\right)=\left(\begin{array}{rl}20r-2& 0\\ 0& -1\end{array}\right)\left(\begin{array}{r}20l{y}_1(t)\\ y_2(t)\end{array}\right)$

Step-by-step solution

System of Differential Equations

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.

Calculation for decoupled system

From Exercise 6,

$\begin{array}{r}lA=\left(\begin{array}{rl}20l1& -2\\ 3& -4\end{array}\right)\\ v_1=\left(\begin{array}{c}20l2\\ 3\end{array}\right)\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{v}}_2=\left(\begin{array}{c}20l1\\ 1\end{array}\right)\end{array}$

With eigenvalues$-2$, and$-1$.

In order to decouple the equation$x^{\prime }=Ax$, such that,$A=PD{P}^{-1}$and$D=P^{-1}AP$, consider$P=\left(\begin{array}{rl}20l{v}_1& v_2\end{array}\right)=\left(\begin{array}{rl}20l2& 1\\ 3& 1\end{array}\right)$.

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

Substituting$x(t)=Py(t)$into$x^{\prime }=Ax$, we have:

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

Here,$P$is having constant entries, so:

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

From the above two solutions, we have:

$P\frac{d}{dt}(y)=PDy\left\{\mathrm{a}\mathrm{s}{P}^{-1}\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{s}\to y^{\prime }=Dy\right\}$

Hence, the solution is $\left(\begin{array}{r}20l{y}_1^{\prime }(t)\\ y_2^{\prime }(t)\end{array}\right)=\left(\begin{array}{rl}20r-2& 0\\ 0& -1\end{array}\right)\left(\begin{array}{r}20l{y}_1(t)\\ y_2(t)\end{array}\right)$.