If we have $\displaystyle \sum_{i=0}^{n} A_i \dfrac{d^i y}{d x^i} =0$, why does the soloution have the form $\displaystyle e^{z x}$ as suggested by Euler. Is there any proof to that?

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- May 24th 2009, 03:30 AMfobos3Linear Differential Homogeneous Equations
If we have $\displaystyle \sum_{i=0}^{n} A_i \dfrac{d^i y}{d x^i} =0$, why does the soloution have the form $\displaystyle e^{z x}$ as suggested by Euler. Is there any proof to that?

- May 24th 2009, 07:35 AMHallsofIvy
It

**doesn't**. For example, the equation y"= 0 is of that form with n= 2, $\displaystyle A_2= 1$, $\displaystyle A_1= A_0= 2$ but its general solution is y(x)= Cx+ D, not an exponential at all.

Also, y"+ y= 0 , with n= 2, $\displaystyle A_2= A_0= 1$ and $\displaystyle A_1= 0$ has general solution y(x)= C cos(x)+ D sin(x).

You can see that those are solutions by finding the first and second derivatives and plugging them into the equations.

The basic idea is good: by**trying**a solution of the form $\displaystyle e^{zx}$, you get the "characteristic equation" which will then lead you to find the general solution even if it**isn't**exponential. You just need to learn more.

For example, if we put $\displaystyle y= e^{zx}$, $\displaystyle y'= ze^{zx}$, $\displaystyle y"= z^2e^{zx}$ into the two equations above, you get $\displaystyle z^2e^{zx}= 0$ and $\displaystyle z^2e^{zx}+ e^{zx}= 0$. Since $\displaystyle e^{zx}$ is never 0, those reduce to $\displaystyle z^2= 0$ and $\displaystyle z^2+ 1$, respectively.

The first equation gives us z= 0 which tells us $\displaystyle e^{0z}= e^0= 1$ is a solution which is why that constant, D, above, is a solution. Since that is the only root, we must learn other reasons to know that multiplying by x also gives a solution.

The second equation has**no**real solutions- its roots are the imaginary numbers i and -i. You need to learn that "complex exponentials" can be written in terms of sine and cosine.

I suspect you will be learning those things shortly. Have fun! - May 24th 2009, 07:59 AMfobos3
I understand the above. The idea is to use the principle of superposition $\displaystyle y=\sum_{i=1}^n c_i y_i$, where $\displaystyle y_x$ is an independent solution. By substituting $\displaystyle y_x=e^{zx}$ you get a polynomial eqation of degree n, which leads to n roots. But how do you know that the diff. eqn. has n independent roots.

What I mean is: if you apply the substitution you get n independent roots and you apply the principle of superposition. But how do you know that the diff. eqn. has exactly n independent roots i.e. you have found all independent solutions. It can have another root you could be missing.

P.S.

$\displaystyle z$ is a complex number