Question

Prove that the general solution of the homogeneous linear system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{0}& \mathbf{6}& \mathbf{0}\\ \mathbf{1}& \mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathit{X}$ on the interval$(-\infty ,\infty )$is

$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{\left(}\begin{array}{c}\mathbf{6}\\ \mathbf{-}\mathbf{1}\\ \mathbf{5}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{3}\\ \mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}\mathbf{\left(}\begin{array}{c}\mathbf{2}\\ \mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathit{e}}^{\mathbf{3}\mathbf{t}}$

Short answer

Yes, the general solution of the given homogeneous linear system is proved to be $X=c_1\left(\begin{array}{c}6\\ -1\\ 5\end{array}\right)e^{\mathrm{t}}+c_2\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{2\mathrm{t}}+c_3\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}}$ on the interval $\left(-\infty ,\infty \right)$.

Step-by-step solution

Definition of general solution - Homogeneous Systems

Let X_p be a given solution of the non-homogeneous system on an interval, I , then

${\mathit{X}}_{\mathbf{c}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathbf{X}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathbf{X}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathbf{X}}_{\mathbf{n}}$

Denote the general solution on the same interval of the homogeneous system.

We are given,

$X\text{'}=\left(\begin{array}{ccc}0& 6& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)X$ ...(1)

Where,

localid="1663913484763" $A=\left(\begin{array}{ccc}0& 6& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$

So, we have the form $X^{\text{'}}=AX$

And,

localid="1663912821685" $X=c_1\left(\begin{array}{c}6\\ -1\\ 5\end{array}\right)e^{\mathrm{t}}+c_2\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{2\mathrm{t}}+c_3\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}}$ ...(2)

Obtaining the derivative of given vector

From equation (2),

$X=c_1\left(\begin{array}{c}6\\ -1\\ 5\end{array}\right)e^{\mathrm{t}}+c_2\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{2\mathrm{t}}+c_3\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}}$

We get,

$X_1=\left(\begin{array}{c}6\\ -1\\ 5\end{array}\right)e^{\mathrm{t}}$

$X_2=\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{2\mathrm{t}}$

$X_3=\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}}$

So, find the derivative of $X_1,X_2,X_3$

role="math" localid="1663913423967" $X_1^{\text{'}}=\left(\begin{array}{c}-6\\ 1\\ 5\end{array}\right)e^{-\mathrm{t}}$

$X_2^{\text{'}}=\left(\begin{array}{c}6\\ -2\\ -2\end{array}\right)e^{-2\mathrm{t}}$

$X_3^{\text{'}}=\left(\begin{array}{c}6\\ 3\\ 3\end{array}\right)e^{3\mathrm{t}}$

Now, ${X_1}^{\text{'}}=A{X}_1$

role="math" localid="1663913757042" $$\begin{gathered} \left(\begin{array}{c}-6\\ 1\\ 5\end{array}\right)e^{-\mathrm{t}}=\left(\begin{array}{ccc}0& 6& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}6\\ -1\\ -5\end{array}\right)e^{\mathrm{t}} \\ \left(\begin{array}{c}-6\\ 1\\ 5\end{array}\right)e^{-\mathrm{t}}=\left(\begin{array}{c}0-6-0\\ 6-0-5\\ 6-1-0\end{array}\right)e^{-\mathrm{t}} \\ \left(\begin{array}{c}-6\\ 1\\ 5\end{array}\right)e^{-\mathrm{t}}=\left(\begin{array}{c}-6\\ 1\\ 5\end{array}\right)e^{-\mathrm{t}} \end{gathered}$$

Thus, ${X_1}^{\text{'}}=A{X}_1$

Next, ${X_2}^{\text{'}}=A{X}_2$

role="math" localid="1663914095800" $$\begin{gathered} \left(\begin{array}{c}6\\ -2\\ -2\end{array}\right)e^{-2\mathrm{t}}=\left(\begin{array}{ccc}0& 6& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{-2\mathrm{t}} \\ \left(\begin{array}{c}6\\ -2\\ -2\end{array}\right)e^{-2\mathrm{t}}=\left(\begin{array}{c}0+6+0\\ -3+0+1\\ -3+1+0\end{array}\right)e^{-2\mathrm{t}} \\ \left(\begin{array}{c}6\\ -2\\ -2\end{array}\right)e^{-2\mathrm{t}}=\left(\begin{array}{c}6\\ -2\\ -2\end{array}\right)e^{-2\mathrm{t}} \end{gathered}$$

Thus,${X_2}^{\text{'}}=A{X}_2$

Next, ${X_3}^{\text{'}}=A{X}_3$

$$\begin{gathered} \left(\begin{array}{c}6\\ 3\\ 3\end{array}\right)e^{3\mathrm{t}}=\left(\begin{array}{ccc}0& 6& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}} \\ \left(\begin{array}{c}6\\ 3\\ 3\end{array}\right)e^{3\mathrm{t}}=\left(\begin{array}{c}0+6+0\\ 2+0+1\\ 2+1+0\end{array}\right)e^{3t} \\ \left(\begin{array}{c}6\\ 3\\ 3\end{array}\right)e^{3\mathrm{t}}=\left(\begin{array}{c}6\\ 3\\ 3\end{array}\right)e^{3\mathrm{t}} \end{gathered}$$

Thus, ${X_3}^{\text{'}}=A{X}_3$

Therefore, X₁, X₂ and X₃ satisfies the given system.

Finding fundamental set of solutions

Find out, if X₁, X₂ and X₃ form a fundamental set of solutions on

$$\begin{gathered} W\left(X_1,X_2,X_3\right)=\left|\begin{array}{ccc}6e^{-t}& -3e^{-2t}& 2e^{3t}\\ -1e^{-t}& 1e^{-2t}& 1e^{3t}\\ -5e^{-t}& 1e^{-2t}& 1e^{3t}\end{array}\right| \\ =6e^{-t}\left[e^{-2t}{e}^{3t}-e^{3t}{e}^{-2t}\right]+3e^{-2t}\left[-e^{-t}{e}^{3t}+5e^{-t}{e}^{3t}\right]+2e^{3t}\left[-e^{-t}{e}^{-2t}+5e^{-t}{e}^{-2t}\right] \\ =3e^{-2t}\left[4e^{2t}\right]+2e^{3t}\left[4e^{-3t}\right] \\ =12+8 \\ =20 \end{gathered}$$

$W\left({\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3\right)\ne 0$

Thus, X₁, X₂ and X₃ forms a fundamental set of solutions i.e., linearly independent on the interval

Therefore, the general solution of the given homogeneous linear system is proved to be on the interval role="math" localid="1663915147769" $\left(-\infty ,\infty \right)$.

Therefore, the general solution of the given homogeneous linear system is proved to be $X=c_1\left(\begin{array}{c}6\\ -1\\ 5\end{array}\right)e^{\mathrm{t}}+c_2\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)e^{2\mathrm{t}}+c_3\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)e^{3\mathrm{t}}$ on the interval $(-\infty ,\infty )$.