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## Homework Statement

Why does the following ODE ALWAYS have two linearly independent solutions?

x''(t) + a(t) x'(t) + b(t) x(t) = f(t)

The characteristic polynomial argument is not sufficient?

- Thread starter LFCFAN
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- #1

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Why does the following ODE ALWAYS have two linearly independent solutions?

x''(t) + a(t) x'(t) + b(t) x(t) = f(t)

The characteristic polynomial argument is not sufficient?

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HallsofIvy

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Any mathematical elaboration I can have?

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HallsofIvy

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In order to show that a set, with a given "addition" and "scalar multiplication" is a vector space you must show three things:

1) It is non empty. You can do that by showing that it contains the "zero vector" (the additive identity). Here, since the addition is ordinary addition of functions, the additive identity is the 0 function: f(x)= 0 for all x. Show that y= 0 satisfies [tex]a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0[/tex] for any functions. (The superscript in parentheses indicates the derivative.)

2) It is "closed under scalar multiplication". To do that show that if y(t) satisfies [tex]a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0[/tex] then so does ay(t) for any number a. That's just a matter of factoring out a.

3) It is "closed under addition". If y= u(t) and y= v(t) both satisfy [tex]a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0[/tex] then so does y= u+ v. That can be done because the nth derivative is itself "linear"- the nth deriviative of u+ v is the nth derivative of u plus the nth derivative of v.

To show that the solution set of a linear homogeneous nth order equation is a vector space of dimension n, use the "existence and uniqueness theorem" to define the n functions, [itex]u_1[/itex] that satisfies the given equation with initial conditions [itex]u_1(0)= 1[/itex], [itex]u_1'(0)= u_1''(0)= \cdot\cdot\cdot= u_1^{(n-1)}(0)= 0[/itex], [itex]u_2(0)= 0[/itex], [itex]u_2'(0)= 1[/itex], [itex]u_2''(0)= \cdot\cdot\cdot= u_2^{(n-1)}(0)= 0[/itex], etc., with "[itex]u_i(t)[/itex]" being the solution to the differential equation having its ith derivative equal to 1 at t= 0 and all other values 0.

Those are sometimes referred to as the "fundamental solutions" to the differential equation. They have this very nice property: if y(t) is

Now, for the non-homogeneous equation, let

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