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

Aug 14, 2017

Yes $F \left(x\right) = G \left(x\right)$

#### Explanation:

We have :

$y = F \left(x\right)$ and $y = G \left(x\right)$ both satisfy $\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right)$ and $y \left({x}_{0}\right) = {y}_{0}$

Consider $y = F \left(x\right)$

It satisfies the equation:

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

Hence $F \left(x\right)$ is an antiderivative of $f \left(x\right)$ and by the FTC, we have:

${y}_{f} = \int \setminus f \left(x\right) \setminus \mathrm{dx}$
${y}_{f} = F \left(x\right) + A$

Using an identical argument for $G \left(x\right)$ we also have:

${y}_{g} = G \left(x\right) + B$

And using the initial condition

$y \left({x}_{0}\right) = {y}_{0} \implies F \left({x}_{0}\right) = G \left({x}_{0}\right) = {y}_{0}$

So we have:

${y}_{0} = F \left({x}_{0}\right) + A = {y}_{0} + A \implies A = 0$
${y}_{0} = G \left({x}_{0}\right) + B = {y}_{0} + B \implies B = 0$

And we therefore conclude that

${y}_{f} = {y}_{g} \implies F \left(x\right) = G \left(x\right)$

ie the solution to a First Order Differential Equation is Unique

Aug 14, 2017

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 $\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right)$ with $y \left({x}_{0}\right) = {y}_{0}$ 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 $F \left(x\right) = G \left(x\right)$

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

${\int}_{x} _ {0}^{x} f \left(x\right) \mathrm{dx} = F \left(x\right) - F \left({x}_{0}\right) = G \left(x\right) - G \left({x}_{0}\right)$

Knowing that $F \left({x}_{0}\right) = g \left({x}_{0}\right)$ allows us to conclude that $F \left(x\right) = G \left(x\right)$.

So the solution is unique.

Aug 14, 2017

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 $F \left(x\right) = G \left(x\right)$ necessarily, if they both solve the same IVP. We know the solution exists, but is $F \left(x\right)$ the only solution?

$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) , \text{ ""I.C.": " } y \left({x}_{0}\right) = {y}_{0}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dF}}{\mathrm{dx}} = f \left(x\right) , \text{ ""I.C.": " } y \left({x}_{0}\right) = {y}_{0}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dG}}{\mathrm{dx}} = f \left(x\right) , \text{ ""I.C.": " } y \left({x}_{0}\right) = {y}_{0}$

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

For the moment, we assume that $F \left(x\right) \ne G \left(x\right)$.

$\int \mathrm{dF} = \int f \left(x\right) \mathrm{dx} = F \left(x\right) + {C}_{1}$

$\int \mathrm{dG} = \int f \left(x\right) \mathrm{dx} = G \left(x\right) + {C}_{2}$

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:

$y \left({x}_{0}\right) = {y}_{0} = F \left({x}_{0}\right) = G \left({x}_{0}\right)$

And we have

${y}_{0} = F \left({x}_{0}\right) + {C}_{1}$

${y}_{0} = G \left({x}_{0}\right) + {C}_{2}$

And $\cancel{{y}_{0}} = \cancel{{y}_{0}} + {C}_{1} = \cancel{{y}_{0}} + {C}_{2}$. Since ${C}_{1} = {C}_{2}$, the solution is unique and $F = G$.