It

**doesn't**. For example, the equation y"= 0 is of that form with n= 2,

,

but its general solution is y(x)= Cx+ D, not an exponential at all.

Also, y"+ y= 0 , with n= 2,

and

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

, 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

,

,

into the two equations above, you get

and

. Since

is never 0, those reduce to

and

, respectively.

The first equation gives us z= 0 which tells us

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!