# Find solution: $\frac {d^4x}{dt^4}=x$

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.

• right, have changed it, thanks May 15 '21 at 18:50
• Please recheck $x',x'''$, although that error does Not appear in your vector $x_0$ May 15 '21 at 18:56
• Ok i see it, there should be $x'=A_1e^t-A_2e^{-t}-A_3\sin{t}+A_4\cos{t}$ and $x'''(t)=A_1e^t-A_2e^{-t}+A_3\sin{t}-A_4\cos{t}$ May 15 '21 at 19:19
• In English we say "bounded" not "limited". May 15 '21 at 19:20

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*}