2nd order homogenous differential equation

[zanazzi78]

Solve the following for y(x);

y'' - 3y' + 2y = 0

I kind of know what to do up to a point but after that I`m stuck (bad notes and no textbook!).

Here`s what i`ve done so far, if someone could hint as how to finish this question i should be able to do the other 9 I have.


let y = e^rx then y' = r e^rx and y''= r^2 e^rx

therefor

r^2 e^rx -3re^rx + 2e^rx = 0

or
r^2 -3r +2 = 0

or

(r-2)(r-1)=0

i.e. r=2 and r=1


now i`m stuck! What do i need to do to find the solution?

(sorry about the lack of latex but my pc refuses to display it! so if you could not use it i would be gratefull)

 

[TD]

Well, as you said you proposed solutions of the form y = e^{rx} and you found r to be 2 and 1. A lineair combination of these solutions, together with two constants, is your complete solution to the homogenous equation.

 

[zanazzi78]

So i would have a solution something like

y= A e^2x + B e^x

How do I find A and B or is that an irrelavent question?

 

[TD]

So i would have a solution something like

y= A e^2x + B e^x

How do I find A and B or is that an irrelavent question?

Your solution is fine and you cannot determine A and B, unless there are initial or boundary conditions. For a general n-th order DE, you'll have n constants.

 

[zanazzi78]

TD thank you very much, your help is gratefully received.

 

[TD]

TD thank you very much, your help is gratefully received.

You're welcome