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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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?
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.Quote:
Originally Posted by fobos3
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!
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.Quote:
Originally Posted by HallsofIvy
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
