# Why do 2nd-order linear ODEs have at most two independent solutions?

1. Aug 11, 2014

### LFCFAN

1. The problem statement, all variables and given/known data

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?

2. Aug 11, 2014

### HallsofIvy

Staff Emeritus
That's a linear equation. It's fairly easy to show that the set of all solutions to an nth order linear homogeneous (f(t)= 0) differential equation forms an n dimensional vector space with the usual addition of functions and multiplication by a number. With f(t) NOT 0, the solution set is a "linear manifold" (think of a plane that does NOT contain the origin) so there exist a function such that every solution can be written as a member of an n dimensional vector space plus that function. The two linearly independent solutions are the "basis vectors" of the two dimensional vector space.

3. Aug 11, 2014

### LFCFAN

Any mathematical elaboration I can have?

4. Aug 11, 2014

### HallsofIvy

Staff Emeritus
Do you understand what a "vector space" is?

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 $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ 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 $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ 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 $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ 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, $u_1$ that satisfies the given equation with initial conditions $u_1(0)= 1$, $u_1'(0)= u_1''(0)= \cdot\cdot\cdot= u_1^{(n-1)}(0)= 0$, $u_2(0)= 0$, $u_2'(0)= 1$, $u_2''(0)= \cdot\cdot\cdot= u_2^{(n-1)}(0)= 0$, etc., with "$u_i(t)$" 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 any solution to the differential equation, let $A_1= y(0)$, $A_2= y'(0)$, ... , $A_{n}= y^{(n-1)}(0)$. Then it is easy to show that $y(t)= A_1u_1(t)+ A_2 u_2(t)+ \cdot\cdot\cdot+ A_{n}u_{n-1}(t)$. That is, any solution to the differential equation can be written as a linear combination of the "fundamental solutions" so they form a basis for the vector space of all solutions showing that the vector space has dimension n. Any solution can be written as a linear combination of those n independent solutions.

Now, for the non-homogeneous equation, let Y(t) be any function satisfying the equation $a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= f(x)$. Then it is easy to show that if y is any other function satisfying that equation, u(t)= y(t)-Y(t) satisfies the equation $a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$. That is, that y(t)= u(t)+ Y(t), it is the sum of a function in the vector space of solutions to the homogeneous equation plus that one given solution to the entire equation.