$\frac{\mathrm{d} g(x)}{\mathrm{d}x}=h(x)$ and $\frac{\mathrm{d} h(x)}{\mathrm{d}x}=g(x)$ where $h(x)\neq g(x)$

Question

Is there any other solution to : $$\frac{\mathrm{d} g(x)}{\mathrm{d}x}=h(x)$$ $$\frac{\mathrm{d} h(x)}{\mathrm{d}x}=g(x)$$ other than $h(x)=g(x)=e^x$?

By varying $\alpha,\beta$ in

$$\frac{\mathrm{d} g(x)}{\mathrm{d}x}=\alpha h(x)$$ $$\frac{\mathrm{d} h(x)}{\mathrm{d}x}=\beta g(x)$$

is it possible to obtain $(e^x,e^x) , (\sin (x),\cos(x))$ as solutions when $\alpha = 1, \beta=1$ and $\alpha = 1, \beta=-1$ (without invoking complex analysis) is there any explanation for relationships between $\alpha,\beta$ yielding relationships between $e^x,\sin(x),\cos(x)$?

Answer 1

The system of differential equations can be written in matrix form as $$\frac{d\vec{u}}{dx}=A\vec{u},$$ where $$A=\begin{bmatrix}0&\alpha\\\beta&0\end{bmatrix}\text{ and } \vec{u}=\begin{bmatrix}g(x)\\h(x)\end{bmatrix}.$$ The general solution can then be expressed in terms of the eigenvalues and eigenvectors of $A$.

The eigenvalues are $\lambda_{1,2}=\pm\sqrt{\alpha\beta}$ with corresponding eigenvectors $$\vec{v}_{1,2}=\begin{bmatrix}-\alpha\\\pm\sqrt{\alpha\beta}\end{bmatrix}.$$

The general form of the solution is then given by $$ \vec{u}(x)=c_1\vec{v}_1e^{\lambda_1 x}+c_2\vec{v}_2e^{\lambda_2 x}.$$

In terms of non-exponential solutions, these can be obtained with complex eigenvalues, i.e. $\alpha\beta<0$.

Answer 2

$$\frac{\mathrm{d} g(x)}{\mathrm{d}x}=\alpha h(x),\frac{\mathrm{d} h(x)}{\mathrm{d}x}=\beta g(x)\implies \frac{\mathrm{d^2} g(x)}{\mathrm{d}{x^2}}=\alpha\frac{\mathrm{d} h(x)}{\mathrm{d}x}=\alpha\beta g(x)$$

Let $g(x)=Ae^{at}\implies \frac{\mathrm{d} g(x)}{\mathrm{d}x}=Aae^{at}$ and $ \frac{\mathrm{d^2} g(x)}{\mathrm{d}{x^2}}=Aa^2e^{at}$

So, $$Ae^{at}\alpha\beta= Aa^2e^{at}$$

As $Ae^{at}\ne 0$ for non-trivial solutions, $a^2=\alpha\beta$

So, $g(x)=A_1e^{a_1t}+A_2e^{a_2t}$ where $A_1,A_2$ are arbitrary for constants and $a_1,a_2$ are the roots of $a^2=\alpha\beta$.

If $\alpha=\beta=1, g(x)=A_1e^t+A_2e^{-t}$ as $a^2=1$

If $\alpha=1,\beta=-1;a^2=-1,a=\pm i$ so $g(x)=A_1e^{it}+A_2e^{-it}$ $=(A_1+A_2)\cos t+i(A_1-A_2)\sin t$ using Euler identity.