Question

If `F(x)` and `G(x)` both solve the same initial value problem, then is it true that `F(x) = G(x)`?

Answer 1

Yes `F(x)=G(x)`

Explanation:

We have :

`y=F(x)` and `y=G(x)` both satisfy `dy/dx=f(x)` and `y(x_0)=y_0`

Consider `y=F(x)`

It satisfies the equation:

`dy/dx=f(x) => d/dx( F(x) = f(x)`

Hence `F(x)` is an antiderivative of `f(x)` and by the FTC, we have:

`y_f = int \ f(x) \ dx`
`y_f = F(x) + A`

Using an identical argument for `G(x)` we also have:

`y_g = G(x) + B`

And using the initial condition

`y(x_0)=y_0 => F(x_0)=G(x_0) = y_0`

So we have:

`y_0 = F(x_0) + A = y_0 + A => A = 0`
`y_0 = G(x_0) + B = y_0 + B => B = 0`

And we therefore conclude that

`y_f=y_g => F(x) = G(x)`

ie the solution to a First Order Differential Equation is Unique

Answer 2

Please see below for a discussion of "uniqueness of solutions". and an answer to the question.

Explanation:

The question can be rephrased as:

Consider the initial value problem `dy/dx = f(x)` with `y(x_0) =y_0` on interval `I`.

Must the solution be unique? That is, if `F` and `G` are solutions on `I`, must it be true that `F=G` on `I`?

or

Can there be more that one solution? (This would be non-unique solutions.)

So the phrase Uniqueness of solutions is a heading telling us what the question is about. It is not a part of the question.

Here is my preferred proof the `F(x) = G(x)`

For any `x` in `I`, the Fundamental Theorem of Calculus tells us that

`int_x_0^x f(x) dx = F(x) - F(x_0) = G(x)-G(x_0)`

Knowing that `F(x_0)=g(x_0)` allows us to conclude that `F(x) = G(x)`.

So the solution is unique.

Answer 3

This is asking, "can there be two solutions to the same first-order differential equation?" Every time we formulate a differential equation, we should ask ourselves,

• Does the solution exist?
• Is the solution unique?
• Is the solution continuous with the problem data?

So, you are asked to prove whether or not `F(x) = G(x)` necessarily, if they both solve the same IVP. We know the solution exists, but is `F(x)` the only solution?

`(dy)/(dx) = f(x), " ""I.C.": " "y(x_0) = y_0`

Given that `F(x)` and `G(x)` are both solutions (not necessarily the same functions), we know we can construct two more equations:

`(dy)/(dx) = (dF)/(dx) = f(x), " ""I.C.": " "y(x_0) = y_0`

`(dy)/(dx) = (dG)/(dx) = f(x), " ""I.C.": " "y(x_0) = y_0`

on an interval `I`, because `F` and `G` are both solutions in `I`.

For the moment, we assume that `F(x) ne G(x)`.

`int dF = int f(x)dx = F(x) + C_1`

`int dG = int f(x)dx = G(x) + C_2`

Their arbitrary constants don't necessarily have to be the same, so we suppose they are not for now. They both are given the same initial condition:

`y(x_0) = y_0 = F(x_0) = G(x_0)`

And we have

`y_0 = F(x_0) + C_1`

`y_0 = G(x_0) + C_2`

And `cancel(y_0) = cancel(y_0) + C_1 = cancel(y_0) + C_2`. Since `C_1 = C_2`, the solution is unique and `F = G`.