Nth order linear ode, why do we have n general solutions?

popopopd · Jun 5, 2015

hi, I looked up the existence and uniqueness of nth order linear ode and I grasped the idea of them, but still kind of confused why we get n numbers of general solutions.

SteamKing · Jun 5, 2015

What does the characteristic equation for the ODE tell you?

http://en.wikipedia.org/wiki/Characteristic_equation_(calculus)

HallsofIvy · Jun 7, 2015

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y'(0)= 1, y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y'(0)= 0, y''(0)= 1, y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y'(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

popopopd · Jun 8, 2015

HallsofIvy said:

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y'(0)= 1, y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y'(0)= 0, y''(0)= 1, y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y'(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

Thanks! it helped a lot!

popopopd · Jun 17, 2015

HallsofIvy said:

A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, a_1y_1+ a_2y_2 is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

To prove that, consider the differential equation a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0.
with the following initial values:

I) y(0)= 1, y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, y_0(x), satisfying the differential equation and those initial conditions.

II) y(0)= 0, y'(0)= 1, y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_1(x), satisfying the differential equation and those initial conditions.

III) y(0)= y'(0)= 0, y''(0)= 1, y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0
Again, there exist a unique solution, y_2(x), satisfying the differential equation and those initial conditions.

Continue that, shifting the "= 1" through the derivatives until we get to
X) y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0, y^{(n-1)}(0)= 1.

First, any solution to the differential equation can written as a linear combination of those n solutions:
Suppose y(x) is a solution to the differential equation. Let A_0= y(0), A_1= y'(0), etc. until A_{n-1}= y^{(n-1)}(0).
Then y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x). That can be shown by evaluating both sides, and their derivatives, at x= 0.

Further, that set of n solutions are independent. To see that suppose that, for some numbers, A_0, A_1, \cdot\cdot\cdot, A_{n-1}, A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0- that is, is equal to 0 for all x. Taking x= 0 we must have A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0 for all x. Set x= 0 to see that A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each A_i, in turn, is equal to 0.

Nth order linear ode, why do we have n general solutions?

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