Final Answer:
The set of solutions to the differential equation forms a vector space due to the closure properties of vector spaces under addition and scalar multiplication, satisfying the requirements for a subspace within the vector space of functions.
Step-by-step explanation:
To show that the set of solutions to the differential equation a₂(x)y'' + a₁(x)y' + a₀(x)y = 0 forms a vector space, we need to demonstrate that it satisfies the conditions of a subspace within the vector space of functions.
Closure under Addition
Let's consider two solutions y₁(x) and y₂(x) to the given differential equation: a₂(x)y'' + a₁(x)y' + a₀(x)y = 0.
If we have y₁(x) and y₂(x) as solutions, it means that they satisfy the equation:
a₂(x)y₁'' + a₁(x)y₁' + a₀(x)y₁ = 0
a₂(x)y₂'' + a₁(x)y₂' + a₀(x)y₂ = 0
Now, let's consider the function c₁y₁(x) + c₂y₂(x), where c₁ and c₂ are constants. We aim to show that this linear combination is also a solution to the given differential equation.
Consider z(x) = c₁y₁(x) + c₂y₂(x):
z'' = c₁y₁'' + c₂y₂''
z' = c₁y₁' + c₂y₂'
Substituting these into the differential equation:
a₂(x)z'' + a₁(x)z' + a₀(x)z = a₂(x)(c₁y₁'' + c₂y₂'') + a₁(x)(c₁y₁' + c₂y₂') + a₀(x)(c₁y₁+ c₂y₂)
Using linearity of differentiation and properties of the solutions y₁(x) and y₂(x), this simplifies to:
c₁(a₁(x)y₁'' + a₁(x)y₁' + a₀(x)y₁) + c₂(a₂(x)y₂'' + a₁(x)y₂' + a₀(x)y₂)
Both terms in the parentheses are zero, as y₁(x) and y₂(x) are solutions to the differential equation. Therefore, z(x) = c₁y₁(x) + c₂y₂(x) is also a solution, showing closure under addition.
Closure under Scalar Multiplication
Consider a solution y(x) to the differential equation. We aim to show that (c . y(x) (where c is a constant) is also a solution.
Let z(x) = c . y(x):
z'' = c .t y''
z' = c . y'
Substituting these into the differential equation:
a₂(x)z'' + a₁(x)z' + a₀(x)z = a₂(x)(c.y'') + a₁(x)(c.y') + a₀(x)(c . y)
Factorizing c out:
c(a₂(x)y'' + a₁(x)y' + a₀(x)y) = c . 0 = 0
Since y(x) satisfies the differential equation, c . y(x) also satisfies it, demonstrating closure under scalar multiplication.
Therefore, since the set of solutions to the differential equation satisfies closure under addition and scalar multiplication, it forms a subspace within the vector space of functions, proving that it is a vector space.
Complete Question
Consider the differential equation
a₂(x)y'' + a₁(x)y' + a₀(x)y = 0
Show that the set of solutions to equation forms a vector space.
Hint: Since the solutions are functions, it is enough to show that the set of solutions to equation is a subspace of the vector space of functions.