Differential Equation's: A popular solving method

[AbedeuS]

Hey, I've recently enrolled in a maths module involving a lot of differential equations, most of the material is all fine and dandy, but I get somewhat confused at this method of solving them.

Scenario

Solve:

$$\frac{d^{2}y}{dx{2}} + 2\frac{dy}{dx} + 3y$$

So then there's this technique we were taught to use, where we define the initial function as just $$e^{mx}$$ then differentiate and substitute and stuff, as such:

$$\frac{d^{2}e^{mx}}{dx^{2}}+2\frac{d e^{mx}}{dx}+3e^{mx}$$

To give:
$$m^{2}e^{mx}+2me^{mx}+3e^{mx}$$

Then:
$$e^{mx}(m^{2}+2m+3)$$ and so on and so fourth through quadratic equations and such.

But what am I technically doing, am i just "Guessing" that the origional function was an exponential with a coefficiant on the power, rather than the "True" form of the function instead of just differentiating the origional function?

Or is this method usually employed when the initial function is not known (or is too complex) and the exponential method is just a nice way of dealing with it? Does it always work?

Thanks
-Abe

 

[Mark44]

Hey, I've recently enrolled in a maths module involving a lot of differential equations, most of the material is all fine and dandy, but I get somewhat confused at this method of solving them.

Scenario

Solve:

$$\frac{d^{2}y}{dx{2}} + 2\frac{dy}{dx} + 3y$$

Technically speaking, there's nothing to solve here -- this isn't an equation.
Presumably you want to solve $\frac{d^{2}y}{dx{2}} + 2\frac{dy}{dx} + 3y = 0$

So then there's this technique we were taught to use, where we define the initial function as just $$e^{mx}$$ then differentiate and substitute and stuff, as such:

$$\frac{d^{2}e^{mx}}{dx^{2}}+2\frac{d e^{mx}}{dx}+3e^{mx}$$

To give:
$$m^{2}e^{mx}+2me^{mx}+3e^{mx}$$

Then:
$$e^{mx}(m^{2}+2m+3)$$ and so on and so fourth through quadratic equations and such.

But what am I technically doing, am i just "Guessing" that the origional function was an exponential with a coefficiant on the power, rather than the "True" form of the function instead of just differentiating the origional function?

Yes, you're taking an educated guess that the solution is an exponential function of the form $y = e^{mx}$, and seeing where that guess leads you.
The whole point of solving differential equations is that you don't know the function that satisfies the diff. equation, so you can't just differentiate the "original" function" -- you don't know what it is.

Or is this method usually employed when the initial function is not known (or is too complex) and the exponential method is just a nice way of dealing with it? Does it always work?

Again, you never know what the solution function is.