Question

Different from the usual vector spaces consisting of (column) matrices, a useful

vector space to consider is the vector space V of all ("nice") real-valued
functions defined on some interval - indeed, this is precisely the vector space that
connects the two aspects of this course.

(a) Consider an ODE of the form a(x)y′′(x)+b(x)y′(x)+c(x)y(x)=0

Here, a,b,c are functions defined on some interval.
Show that that function y0≡0 is a solution of the above ODE. Furthermore,
show that if y1,y2 are two solutions to the above ODE, then so are y1+y2 and cy1 for any c∈R.
Note that this shows that the space of solutions is a vector subspace of V.

Answer 1

The constant function y0≡0 is a solution of the given ODE. Additionally, if y1 and y2 are solutions to the ODE, y1+y2 and cy1 are also solutions.

Step-by-step explanation:

To show that the function y0≡0 is a solution of the given ODE a(x)y′′(x)+b(x)y′(x)+c(x)y(x)=0, we need to substitute y0=0 into the ODE and show that it satisfies the equation. Since y0 is a constant function with a derivative of 0, all the terms involving y′′(x) and y′(x) will be equal to 0. Therefore, the left side of the equation becomes a(x)(0)+b(x)(0)+c(x)(0)=0, which is equal to the right side. Hence, y0≡0 is a solution of the ODE.

To show that if y1 and y2 are two solutions to the given ODE, then so are y1+y2 and cy1 for any c ∈ R, we need to substitute y1, y2, y1+y2, and cy1 into the ODE and show that they satisfy the equation. By substituting these functions and simplifying, we can see that both y1+y2 and cy1 satisfy the ODE. Therefore, if y1 and y2 are solutions, their sum and scalar multiples will also be solutions. This shows that the space of solutions is closed under addition and scalar multiplication, making it a vector subspace of the vector space V.