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Find solution: $\frac {d^4x}{dt^4}=x \tag{1}\label{eq1}$ For which $x_0 = (x(0),x'(0),x''(0),x^{(3)}(0))$ the solution is:

  • limited on $(0,\infty)$
  • limited on $(-\infty,0)$
  • periodic

so we have $P(\lambda)=\lambda^4-1$ $$\lambda_1=1,\lambda_2=-1,\lambda_3=i,\lambda_4=-i$$

each funtion $e^{it},e^{-it}$ is a solution to \eqref{eq1} so their linear combination also is a solution to \eqref{eq1}

So we have: $$x(t)=A_1e^t+A_2e^{-t}+A_3\cos{t}+A_4\sin{t}$$

$x'(t)=A_1e^t-A_2e^{-t}-A_3\cos{t}+A_4\sin{t}$

$x''(t)=A_1e^t+A_2e^{-t}-A_3\cos{t}-A_4\sin{t}$

$x^{(3)}(t)=A_1e^t-A_2e^{-t}+A_3\cos{t}-A_4\sin{t}$

So: $$x_0=(A_1+A_2+A_3, A_1-A_2+A_4, A_1+A_2-A_3, A_1-A_2-A_4)$$

Thus the solution is limited on $(0,+\infty) \Leftrightarrow A_1=0$ because then there is no $e^t$ and $\lim_{t \to \infty} e^{-t} = 0$

the solution is limited on $(-\infty,0) \Leftrightarrow A_2=0$, as $\lim_{t \to -\infty} e^{t} = 0$, $\lim_{t \to 0^-} e^{t} = 1$

the solution is periodic:

  • on $(0,\infty) \Leftrightarrow A_1=0$
  • on $(-\infty,0)\Leftrightarrow A_2=0$
  • on $\mathbb{R} \Leftrightarrow A_1=A_2=0$

It is a good solution? If there are some kind of mistakes can I get hints what should I correct, thanks.

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  • $\begingroup$ right, have changed it, thanks $\endgroup$
    – frugo
    May 15 '21 at 18:50
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    $\begingroup$ Please recheck $x',x'''$, although that error does Not appear in your vector $x_0$ $\endgroup$ May 15 '21 at 18:56
  • $\begingroup$ Ok i see it, there should be $x'=A_1e^t-A_2e^{-t}-A_3\sin{t}+A_4\cos{t}$ and $x'''(t)=A_1e^t-A_2e^{-t}+A_3\sin{t}-A_4\cos{t}$ $\endgroup$
    – frugo
    May 15 '21 at 19:19
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    $\begingroup$ In English we say "bounded" not "limited". $\endgroup$
    – GEdgar
    May 15 '21 at 19:20
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Your solution is not complete. You are required to find the admissible vectors $(x(0),x'(0),x''(0),x'''(0))$ but you have found the answers in terms of $A_i$ which are not recognized by the question. Basically you have to express $A_i$ in terms of $x^{(i)}(0)$ knowing that$$\begin{align*}x(0)&=A_1+A_2+A_3\\x'(0)&=A_1-A_2+A_4\\x''(0)&=A_1+A_2-A_3\\x'''(0)&=A_1-A_2-A_4\end{align*}\iff\begin{bmatrix}1&1&1&0&|&x(0)\\1&-1&0&1&|&x'(0)\\1&1&-1&0&|&x''(0)\\1&-1&0&-1&|&x'''(0)\end{bmatrix}\\ \sim\begin{bmatrix}1&0&0&0&|&\frac{x(0)+x'(0)+x''(0)+x'''(0)}4\\0&-2&0&0&|&\frac{x'(0)+x'''(0)-x(0)-x''(0)}2\\0&0&-2&0&|&x''(0)-x(0)\\0&0&0&-2&|&x'''(0)-x'(0)\end{bmatrix}$$giving$$\begin{align*}A_4&=\frac{x'(0)-x'''(0)}2\\A_3&=\frac{x(0)-x''(0)}2\\A_2&=\frac{x(0)+x''(0)-x'(0)-x'''(0)}4\\A_1&=\frac{x(0)+x'(0)+x''(0)+x'''(0)}4\end{align*}$$Now you can answer in terms of $x^{(i)}(0)$:$$\begin{align*}\text{Bounded on }\Bbb R^+&\iff A_1=0\iff x(0)+x'(0)+x''(0)+x'''(0)=0\\\text{Bounded on }\Bbb R^-&\iff A_2=0\iff x(0)+x''(0)=x'(0)+x'''(0)\\\text{Periodic on }I\subseteq\Bbb R&\iff A_1=A_2=0\iff x(0)+x''(0)=x'(0)+x'''(0)=0\end{align*}$$

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