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

Answer 1

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

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.

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Answer 2

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,

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#.
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Answer 3

Yes #F(x)=G(x)#

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

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Answer 4

If ( F(x) ) and ( G(x) ) both solve the same initial value problem, it is not necessarily true that ( F(x) = G(x) ). While both functions may satisfy the initial condition(s) of the problem and produce identical outputs for a particular input, there could still be differences between the functions themselves. These differences could arise due to various reasons such as the methods used for solving the problem, the domain of definition, or the presence of arbitrary constants in the solutions. Therefore, while both functions yield the same results under the specified initial conditions, they may not be identical functions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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