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

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SUMMARY

The basis of the vector space of solutions to the differential equation (y'' + 5y' - 50y = 0) is defined as {e^(5x), e^(-10x)}. The general solution can be expressed as y(x) = Ae^(5x) + Be^(-10x), where A and B are constants. To form a basis, specific values for A and B must be chosen such that A + B equals one, leading to particular solutions like y1 = s*e^(5x) + (1-s)e^(-10x) and y2 = t*e^(5x) + (1-t)e^(-10x). The book's answer is derived from selecting A=1 and B=0.

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PhysicsKid42
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Apologies, have solved this question.
Answer if useful for anyone:
Basis= {e^(5x),e^(-10x)}.

Homework Statement



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

Homework Equations



-

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:
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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
 

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