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

• popopopd
In summary, the characteristic equation for the ODE tells you how many solutions there are, as well as their properties.
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.

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.

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

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

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 ODEThanks always, Your answers help me a lot!

Last edited:

## 1. Why do we need to use n general solutions for Nth order linear ODEs?

The order of a differential equation refers to the highest derivative present in the equation. For example, a second order linear ODE contains a second derivative. In general, an Nth order linear ODE will have N general solutions, one for each derivative present. This is because each solution corresponds to a different derivative, and by combining them, we can find a complete solution for the entire equation.

## 2. How do we determine the number of general solutions for an Nth order linear ODE?

The number of general solutions for an Nth order linear ODE is equal to the order of the equation. This means that a second order linear ODE will have two general solutions, a third order linear ODE will have three general solutions, and so on.

## 3. Can we find a unique solution for an Nth order linear ODE?

No, in general, we cannot find a unique solution for an Nth order linear ODE. This is because the number of general solutions is equal to the order of the equation, which means that there are multiple solutions that satisfy the equation. To find a unique solution, we need to have initial conditions or boundary conditions.

## 4. How do we choose which general solutions to use for a particular Nth order linear ODE?

The general solutions for an Nth order linear ODE will depend on the specific form of the equation, as well as any initial or boundary conditions that are given. In general, we will need to use all N general solutions to find a complete solution for the equation.

## 5. Can we use fewer or more than N general solutions for an Nth order linear ODE?

No, we cannot use fewer or more than N general solutions for an Nth order linear ODE. This is because each general solution corresponds to a different derivative in the equation, and using any more or fewer solutions will result in a solution that does not satisfy the equation. However, we can combine the N general solutions in different ways to find unique solutions that satisfy certain initial or boundary conditions.

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