Choice of basis

1. Nov 21, 2005

asdf1

for the problem:
y+y=secx

why can't 0 and e^(-1x) be choices for the basis?

2. Nov 21, 2005

HallsofIvy

Be precise! What you mean is "why can't 0 and e-x be a basis for the solution space of the corresponding homogeneous equation".

I am sorely tempted to ask why in the world you think it should be. For one thing, the "0" vector (here, the function that is identically 0) is never a member of a basis- any set of vectors including 0 cannot be independent. But then I also wonder where you got "e-x[/sup}"!
What is the characteristic equation for y"= y= 0? What are its solutions?

3. Nov 22, 2005

asdf1

i figured that if you wanted to solve the homogenous equation, then suppose that y=e^(namda)x
and subsitute that into the equation to find namda which equals 0 and
(-1)...
so i thought that since those two are the solutions to the homogeneous question, then couldn't they be the basis?

4. Nov 22, 2005

matt grime

When did e^{namdax} (lambda?) become zero when namda is zero?

And the qustion still remains: a basis of what?

You haven't solved y''+y=0 correctly anyway. e^0 is constant, and y=const is not a solution of that equation, nor for that matter is e^{-x}

Last edited: Nov 22, 2005
5. Nov 22, 2005

HallsofIvy

First, it's "lambda", not "nambda". Yes, you "suppose that $y= e^{\lambda x}$ and substitute that into the equation to find lamba which equals..." No, it does not equal 0 and -1! I asked before, what is the characteristic equation? What equation do you get when you substitute $y= e^{\lambda x}$?

6. Nov 24, 2005

asdf1

sorry about the spelling! thanks for correcting that!
the characteristic equation is (lambda)^2+(lambda)=0
=> [(lamda)((lamda)+1)]=0
=>lamda=0, -1 ?

7. Nov 24, 2005

matt grime

no it isn't, not unless you made a typo in the original post. you're characteristic equation is for y'' + y', when you wrote y'' + y

8. Nov 24, 2005

HallsofIvy

The differential equation was y"+ y= 0. If $y= e^{\lambda x}$ then $y'= \lambdae^{\lambda x}$ and $y"= \lambda^2e^{\lambda x}$. The characteristic equation is [/itex]\lambda^2+ 1= 0[/itex] which has roots i and -i. A basis for the solution space is {sin x, cos x}.
If the original equation was y"+ y'= 0, then the characteristic equation is y"+ y= 0 which has roots 0 and -1. In that case a basis for the solution space would {e0x= 1, e-x}

9. Nov 25, 2005

asdf1

opps... thank you for correcting me!!!