# Basis of the vector space of solutions to a differential equation.

1. Mar 5, 2009

### PhysicsKid42

Apologies, have solved this question.
Answer if useful for anyone:
Basis= {e^(5x),e^(-10x)}.

1. The problem statement, all variables and given/known data

Consider the differential equation
(2nd derivative of y wrt x) + 5(1st derivative of y wrt x) - 50y =0

Find a basis of the vector space of solutions of the above differential equation. You should separate each vector with a comma, and each should take the value of 1 at x=0

2. Relevant equations

-

3. The attempt at a solution

Solution: y(x)=Ae^(5x) + Be^(-10x)
The solutions are a linear combination of these two terms.
I'm unsure of how to convert this into 'a basis of the vector space of solutions'.

Last edited: Mar 5, 2009
2. Mar 5, 2009

### lurflurf

So take values of A and B, to give particular solutions. List them.
The value must not be the same and A+B must equal one.
say s!=t
y1=s*exp(5x)+(1-s)*exp(-10x)
y2=t*exp(5x)+(1-t)*exp(-10x)

Basis={y1,y2}={s*exp(5x)+(1-s)*exp(-10x),t*exp(5x)+(1-t)*exp(-10x)}
you can also just chose particular s and t
s=10,t=101
Basis={y1,y2}={10*exp(5x)-9*exp(-10x),101*exp(5x)-100*exp(-10x)}
The book answer results from the choice s=1,t=0