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 $e^x,\sin(x),\cos(x)$?

  • $\begingroup$ This is a linear system, so it can be represented in matrix form. $\endgroup$
    – Daryl
    Jan 1, 2013 at 10:30
  • $\begingroup$ @Daryl : Is That matrix form equivalent to using complex form? $\endgroup$
    – jimjim
    Jan 1, 2013 at 10:33

2 Answers 2


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. $\alpha\beta<0$.


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

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.

  • $\begingroup$ I bet this generalizes really nicely to $f=af^{(n)}$. $\endgroup$
    – Rhymoid
    Jan 1, 2013 at 10:44
  • $\begingroup$ Your line after "So, ..." is incorrect. It should be $Aae^{at}=\alpha\beta Ae^{at}$, as you substituted $g$ and $g''$ in the wrong places. $\endgroup$
    – Daryl
    Jan 1, 2013 at 10:50
  • $\begingroup$ @Daryl, I think you meant $Aa^2e^{at}=\alpha\beta Ae^{at}$ as rectified in the answer? $\endgroup$ Jan 1, 2013 at 10:55
  • $\begingroup$ +1, but is there an alternative to Euler identity, or is it a fundamental block of math that can not do without? No alternatives in it's place? $\endgroup$
    – jimjim
    Jan 1, 2013 at 10:55
  • 2
    $\begingroup$ @Arjang, to me this identity is indispensable in (en.wikipedia.org/wiki/Linear_differential_equation) $\endgroup$ Jan 1, 2013 at 11:02

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