# Determining whether this equation is a subspace?

## Homework Statement

There is a vector space with set F, of all real functions. It has the usual operations of addition of functions and multiplication by scalars. You have to determine whether this equation is a subspace of F: $$f''(x) + 3f'(x) + x^2 f(x) = sin(x)$$

## Homework Equations

$$f''(x) + 3f'(x) + x^2 f(x) = sin(x)$$ the 0 vector/function

## The Attempt at a Solution

So, to test that it is non-empty set I used the 0 vector/function. However, is this the same as letting "x=0"? If so, it would then be:
$$f''(0) + 3f'(0) + x^2 f(0) = sin(0)$$ and therefore $$0 = 0$$ proving that the set is non-empty.
As, wouldn't it be what value also makes sin(x) = 0 (which is x=0) and so, would this be correct?
I just want to clarify whether it is before I continue further with solving the problem.

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Fredrik
Staff Emeritus
Gold Member
Equations can't be subspaces. I assume that the problem is asking you to check if the set of solutions of the equation is a subspace of F. Note that this set is a subset of F.

The zero vector in F is the function that takes every x in ℝ to the number 0. You should verify that this function has the properties a zero vector is supposed to have.

Last edited:
HallsofIvy
Homework Helper

## Homework Statement

There is a vector space with set F, of all real functions. It has the usual operations of addition of functions and multiplication by scalars. You have to determine whether this equation is a subspace of F: $$f''(x) + 3f'(x) + x^2 f(x) = sin(x)$$
As Fredrik said, an equation is not a "subspace" what you want to determine is whether the set of all functions satisfying that equation is a subspace.

## Homework Equations

$$f''(x) + 3f'(x) + x^2 f(x) = sin(x)$$ the 0 vector/function

## The Attempt at a Solution

So, to test that it is non-empty set I used the 0 vector/function. However, is this the same as letting "x=0"? If so, it would then be:
No, it is not, the 0 "function" is f(x)= 0 for all x.

$$f''(0) + 3f'(0) + x^2 f(0) = sin(0)$$ and therefore $$0 = 0$$ proving that the set is non-empty.
No, that is incorrect. You cannot set x= 0. If f(x)= 0 for all x, then its first and second derivatives are also 0 but the right hand side is not.

As, wouldn't it be what value also makes sin(x) = 0 (which is x=0) and so, would this be correct?
I just want to clarify whether it is before I continue further with solving the problem.