Are There Multiple Solutions to This ODE Problem?

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The discussion centers on proving the existence of multiple solutions to the ODE dx/dt = √[3]{x²} with the initial condition x(0) = 0. One solution presented is x = (1/3 t)^(1/3), while another solution is x = 0, which satisfies the initial condition. Participants note that the professor's focus on advanced topics may have left gaps in understanding ODE techniques. The importance of separation of variables is highlighted as a method to derive solutions. Both solutions are confirmed as valid, demonstrating the existence of multiple solutions to the problem.
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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?
 
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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