Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
This is a linear system, so it can be represented in matrix form. – Daryl Jan 1 '13 at 10:30
@Daryl : Is That matrix form equivalent to using complex form? – Arjang Jan 1 '13 at 10:33
up vote 2 down vote accepted

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. $\alpha\beta<0$.

share|cite|improve this answer

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

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.

share|cite|improve this answer
I bet this generalizes really nicely to $f=af^{(n)}$. – Rhymoid Jan 1 '13 at 10:44
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. – Daryl Jan 1 '13 at 10:50
@Daryl, I think you meant $Aa^2e^{at}=\alpha\beta Ae^{at}$ as rectified in the answer? – lab bhattacharjee Jan 1 '13 at 10:55
+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? – Arjang Jan 1 '13 at 10:55
@Arjang, to me this identity is indispensable in ( – lab bhattacharjee Jan 1 '13 at 11:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.