Question

Question: In Problems 1-12 find the general solution of the given system.
$$\begin{gathered} \frac{dx}{dt}=2x-7y\backslash \mathrm{hfill} \\ \frac{dy}{dt}=5x+10y+4z \\ \frac{dz}{dt}=5y+2z \end{gathered}$$

Short answer

The general solution of the given system is$$\begin{gathered} \mathrm{X}={\mathrm{c}}_1\left(4 \\ 0 \\ -5\right){\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_2\left(-7 \\ 3 \\ 5\right){\mathrm{e}}^{5\mathrm{t}}+{\mathrm{c}}^3\left(-7 \\ 5 \\ 5\right){\mathrm{e}}^{7\mathrm{t}} \end{gathered}$$

Step-by-step solution

Definition of General Solution of Homogeneous Systems

Let${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$be ndistinct real eigenvalues of the coefficient matrixA

of the homogeneous system and let ${\mathit{K}}_{\mathbf{1}}\mathbf{,}{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{K}}_{\mathbf{n}}$be the corresponding

eigenvectors. Then the general solution on the interval $\left(-\infty ,\infty \right)$is given by$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{K}}_{\mathbf{1}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{K}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathit{K}}_{\mathbf{n}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{n}}\mathbf{t}}$.

The given information is written as:
$$\begin{gathered} \frac{dx}{dt}=2x-7y \\ \frac{dy}{dt}=5x+10y+4z \\ \frac{dz}{dt}=5y+2z \end{gathered}$$
And this can be written in the matrix form as:
$$\begin{gathered} \mathrm{X}=\left(2-70 \\ 5104 \\ 052\right)\mathrm{X}\mathrm{where}\mathrm{A}=\left(2-70 \\ 5104 \\ 022\right) \end{gathered}$$

Eigenvalues of the Coefficient Matrix

Obtaining the eigenvalues of the coefficient matrix as:

So our eigenvalues are ${\lambda }_1=2$, ${\lambda }_2=5$and ${\lambda }_3=7$.

Eigenvector and its Corresponding Solution Vector

For${\lambda }_1=2$, find its corresponding solution vector as:
$$\begin{gathered} \left(2-2-70 \\ 510-24 \\ 052-2\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \\ \left(0-70 \\ 584 \\ 050\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \end{gathered}$$
Therefore, the three equations can be written as:
$-7y=0$ ………….. (1)
$5x+8y+4z=0$ …………. (2)
$5y=0$ …………. (3)

Since $5y=0$, $y=0$.

Since $y=0$,localid="1663948836708" $5x+8y+4z=0 \Rightarrow 5x=-4z$

Choosing $\mathrm{x}=4$yields$z=-5$ .
This gives an eigenvector and a corresponding solution vector $$\begin{gathered} {\mathrm{K}}_1=\left(4 \\ 0 \\ -5\right),{\mathrm{X}}_1=\left(4 \\ 0 \\ -5\right){\mathrm{e}}^{2\mathrm{t}} \end{gathered}$$
For${\lambda }_2=5$, find its corresponding solution vector as:
$$\begin{gathered} \left(2-5-70 \\ 510-54 \\ 052-5\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \\ \left(-3-70 \\ 554 \\ 05-3\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \end{gathered}$$

Therefore, the three equations can be written as:
$-3x-7y=0$ ………….. (4)
$5x+5y+5z=0$ …………. (5)
$5y-3z=0$ …………. (6)

Rewriting equation (4) as:
$$\begin{gathered} -3x=7y \\ x=-\frac{7}{3}y \end{gathered}$$
Choosing $y=3$yields$x=-7$.

Rewriting equation (6) as:
$$\begin{gathered} 5y=3z \\ z=\frac{5}{3}y \end{gathered}$$

Choosing $\mathrm{y}=3$yields$\mathrm{z}=5$ .
This gives an eigenvector and a corresponding solution vector localid="1663948844982" $$\begin{gathered} {\mathrm{K}}_2=\left(-7 \\ 3 \\ 5\right),{\mathrm{X}}_2=\left(-7 \\ 3 \\ 5\right){\mathrm{e}}^{5\mathrm{t}} \end{gathered}$$

For${\lambda }_3=7$, find its corresponding solution vector as:

$$\begin{gathered} \left(2-7-70 \\ 510-74 \\ 052-7\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \\ \left(-5-70 \\ 534 \\ 05-5\right)\left(x \\ y \\ z\right)=\left(0 \\ 0 \\ 0\right) \end{gathered}$$

Therefore, the three equations can be written as:
$-5x-7y=0$ ………….. (7)
$5x+3y+4z=0$…………. (8)

$5y-5z=0$ …………. (9)

Rewriting equation (7) as:

localid="1663943494378" $$\begin{gathered} -5\mathrm{x}=7\mathrm{y} \\ \mathrm{x}=-\frac{7}{5}\mathrm{y} \end{gathered}$$

Choosing localid="1663942720615" $y=5$yields $x=-7$.

Rewriting equation (9) as:

localid="1663943396692" $$\begin{gathered} 5y=5z \\ y=z \end{gathered}$$

Choosing $y=5$yields $z=5$.

This gives an eigenvector and a corresponding solution vector$$\begin{gathered} {\mathrm{K}}_3=\left(-7 \\ 5 \\ 5\right){\mathrm{X}}_3=\left(-7 \\ 5 \\ 5\right){\mathrm{e}}^{7\mathrm{t}} \end{gathered}$$

General solution of given system

Finally, the general solution is found using $$\begin{gathered} {\mathrm{X}}_1=\left(4 \\ 0 \\ -5\right){\mathrm{e}}^{2\mathrm{t}},{\mathrm{X}}_2=\left(-7 \\ 3 \\ 5\right){\mathrm{e}}^{5\mathrm{t}} \end{gathered}$$ and
$$\begin{gathered} {\mathrm{X}}_3=\left(-7 \\ 5 \\ 5\right){\mathrm{e}}^{7\mathrm{t}}\mathrm{as}: \end{gathered}$$

Therefore, the general solution of the given system is