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

3 Answers
Aug 14, 2017

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

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 #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.

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(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#.