Are There Multiple Solutions to This ODE Problem?

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Homework Help Overview

The discussion revolves around proving the existence of multiple solutions to the ordinary differential equation (ODE) given by \(\frac{dx}{dt} = \sqrt[3]{x^{2}}\) with the initial condition \(x(0) = 0\). Participants are exploring the implications of this setup and the nature of the solutions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants present various attempts to find solutions, including expressions for \(x(t)\) based on the initial condition. Some question the correctness of each other's proposed solutions and seek clarification on the reasoning behind them.

Discussion Status

There is an ongoing examination of different proposed solutions, with some participants confirming that \(x=0\) is indeed a valid solution. However, there is no explicit consensus on the correctness of the other solutions presented, and participants are actively engaging in clarifying their approaches.

Contextual Notes

Some participants note that the course content may be more advanced than typical differential equations, suggesting that foundational knowledge in ODEs might be assumed. This context may influence the understanding and interpretation of the problem.

brainslush
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Homework Statement


Proof that there exist more than one solution to following equation
\frac{dx}{dt} = \sqrt[3]{x^{2}} , x(0) = 0

Homework Equations


The Attempt at a Solution


Well, I need a confirmation to my attempt of solution. The one is quite forward:

\Rightarrow x=(1/3(t+c))^{3}

Pluging in x(0) = 0 yields that:

\Rightarrow x=(1/3(t))^{3} is a solution.

The point is, my professor never told us how to solve ODEs. Most of the time he's talking about manifolds, charts, diffeomorphisms, etc.

So I guess the other solution one finds doing following:
Since x(0) = 0 \Rightarrow \dot{x}(0) = 0

Pluging in \dot{x} one gets:

0 = \sqrt[3]{x^{2}}

and therefore the other solution is x = 0

Are my solutions correct?
 
Last edited:
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brainslush said:

Homework Statement


Proof that there exist more than one solution to following equation
\frac{dx}{dt} = \sqrt[3]{x^{2}} , x(0) = 0


Homework Equations





The Attempt at a Solution


Well, I need a confirmation to my attempt of solution. The one is quite forward:

\Rightarrow x=\sqrt[3]{3(t+c)}

I don't see how you got that solution? I don't even think it's correct. Could you elaborate?

Pluging in x(0) = 0 yields that:

\Rightarrow x=\sqrt[3]{3t} is a solution.

The point is, my professor never told us how to solve ODEs. Most of the time he's talking about manifolds, charts, diffeomorphisms, etc.

So I guess the other solution one finds doing following:
Since x(0) = 0 \Rightarrow \dot{x}(0) = 0

Pluging in \dot{x} one gets:

0 = \sqrt[3]{x^{2}}

and therefore the other solution is x = 0

Are my solutions correct?

The x(t)=0 solution is good!
 
Upss my bad, totally messed up the first solution
 
If you don't know it, the trick in finding the solution to such an equation is "separation of the variables".

Say, for example, that you have the ODE

\frac{dx}{dt}=\frac{t}{x}

then you rewrite it as

xdx=tdt

Integrating gives you

\int{xdx}=\int{tdt}

Thus

\frac{x^2}{2}=\frac{t^2}{2}+C

and so we get that

x=\pm\sqrt{t^2+C}

is the answer. Try something analogous for your equation...
 
brainslush said:

Homework Statement


Proof that there exist more than one solution to following equation
\frac{dx}{dt} = \sqrt[3]{x^{2}} , x(0) = 0


Homework Equations





The Attempt at a Solution


Well, I need a confirmation to my attempt of solution. The one is quite forward:

\Rightarrow x=(1/3(t+c))^{3}
This should be to the 1/3 power, not 3.

Pluging in x(0) = 0 yields that:

\Rightarrow x=(1/3(t))^{3} is a solution.
x= ((1/3)t)^{1/3} is a solution.

The point is, my professor never told us how to solve ODEs. Most of the time he's talking about manifolds, charts, diffeomorphisms, etc.
Sounds like the course is much more advanced than a differential equations course so your professor is assuming you have already taken a course in differential equations.

So I guess the other solution one finds doing following:
Since x(0) = 0 \Rightarrow \dot{x}(0) = 0

Pluging in \dot{x} one gets:

0 = \sqrt[3]{x^{2}}

and therefore the other solution is x = 0

Are my solutions correct?
x= 0 is definitely a solution!
 

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