Linear Differential Homogeneous Equations

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May 24th 2009, 03:30 AM
fobos3

Linear Differential Homogeneous Equations

If we have $\sum_{i=0}^{n} A_i \dfrac{d^i y}{d x^i} =0$ , why does the soloution have the form $e^{z x}$ as suggested by Euler. Is there any proof to that?
May 24th 2009, 07:35 AM
HallsofIvy

Quote:

Originally Posted by fobos3

If we have $\sum_{i=0}^{n} A_i \dfrac{d^i y}{d x^i} =0$ , why does the soloution have the form $e^{z x}$ 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, $A_2= 1$ , $A_1= A_0= 2$ but its general solution is y(x)= Cx+ D, not an exponential at all.

Also, y"+ y= 0 , with n= 2, $A_2= A_0= 1$ and $A_1= 0$ 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 $e^{zx}$ , 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 $y= e^{zx}$ , $y'= ze^{zx}$ , $y"= z^2e^{zx}$ into the two equations above, you get $z^2e^{zx}= 0$ and $z^2e^{zx}+ e^{zx}= 0$ . Since $e^{zx}$ is never 0, those reduce to $z^2= 0$ and $z^2+ 1$ , respectively.

The first equation gives us z= 0 which tells us $e^{0z}= e^0= 1$ 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 AM
fobos3

Quote:

Originally Posted by HallsofIvy

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

Also, y"+ y= 0 , with n= 2, $A_2= A_0= 1$ and $A_1= 0$ 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 $e^{zx}$ , 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 $y= e^{zx}$ , $y'= ze^{zx}$ , $y"= z^2e^{zx}$ into the two equations above, you get $z^2e^{zx}= 0$ and $z^2e^{zx}+ e^{zx}= 0$ . Since $e^{zx}$ is never 0, those reduce to $z^2= 0$ and $z^2+ 1$ , respectively.

The first equation gives us z= 0 which tells us $e^{0z}= e^0= 1$ 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 $y=\sum_{i=1}^n c_i y_i$ , where $y_x$ is an independent solution. By substituting $y_x=e^{zx}$ 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.
$z$ is a complex number