Are There Multiple Solutions to This ODE 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?
 
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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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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