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

1. Jun 5, 2015

### popopopd

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.

2. Jun 5, 2015

### SteamKing

Staff Emeritus
3. Jun 7, 2015

### HallsofIvy

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 $A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each [itex]A_i$, in turn, is equal to 0.

4. Jun 8, 2015

### popopopd

Thanks! it helped alot!

5. Jun 17, 2015

### popopopd

(5)
(17)

- using picard's iteration in vector form, to prove nth order linear ODE's existence & uniqueness.

ex
(21)

(22)

(http://ghebook.blogspot.ca/2011/10/differential-equation.html)

Hi, I actually did picard's iteration and found that without n initial conditions, nth order linear ODE will have n number of constants as we assume initial conditions are some arbitrary constants.

since function spaces are vector spaces, solutions span n dimensional vector space (not very sure of this)

If we do picard's iteration,

y(n-1)=y0(n-1)+∫y(n)dx
y(n-2)=y0(n-2)+∫y(n-1)dx
=y(n-2)=y0(n-2)+∫y0(n-1)+∫y(n)dx
.
.
.
iteration goes on and on until the error is sufficiently decreased.

if we assume each initial conditions are some constants, we will eventually sort out the solution function y w.r.t constants after sufficient number of iteration is done. Then it will look like,

y (c0 c1 c2 - - - - - - )[y1 y2 y3 y4 y5 - - - - - - ] <-- (should be vertical)

which is in the form of y = c1y1+c2y2+c3y3 . . .

and so on
is this correct?
If not, how can we show that solution space has n number of basis?

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Also, I have two questions about Strum Liouville 2nd order ODE.

1. if we look at the Strum-Liouville 2nd order ODEs, there is an eigenvalue term within the equation. it seems like we are adding one more constant in the equation, which imposes a restriction to find a solution (n+1 constants with n conditions).

[m(x)y']+[λr(x)-q(x)]y
=m(x)[y''+P(x)y'+Q(x)y]
=m(x)[y''+P(x)y'+(λr(x)-q(x))]
=0

∴ [y''+P(x)y'+(λr(x)-q(x))]=0

if we do Picard's iteration, then we have one more constant λ along with n constants..

2. I don't understand how eigenvalue directly influence the solutions of 2nd order ODE

Thanks always, Your answers help me a lot!

Last edited: Jun 17, 2015