Initial value Problem, ODE

[brainslush]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

Pluging in x(0) = 0 yields that:

[itex]\Rightarrow x=(1/3(t))^{3} [/itex] 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 [itex]x(0) = 0 \Rightarrow \dot{x}(0) = 0[/itex]

Pluging in [itex]\dot{x}[/itex] one gets:

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

and therefore the other solution is x = 0

Are my solutions correct?

 

[micromass]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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:

[itex]\Rightarrow x=\sqrt[3]{3t} [/itex] 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 [itex]x(0) = 0 \Rightarrow \dot{x}(0) = 0[/itex]

Pluging in [itex]\dot{x}[/itex] one gets:

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

and therefore the other solution is x = 0

Are my solutions correct?

The x(t)=0 solution is good!

 

[brainslush]

Upss my bad, totally messed up the first solution

 

[micromass]

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

[tex]\frac{dx}{dt}=\frac{t}{x}[/tex]

then you rewrite it as

[tex]xdx=tdt[/tex]

Integrating gives you

[tex]\int{xdx}=\int{tdt}[/tex]

Thus

[tex]\frac{x^2}{2}=\frac{t^2}{2}+C[/tex]

and so we get that

[tex]x=\pm\sqrt{t^2+C}[/tex]

is the answer. Try something analogous for your equation...

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

This should be to the 1/3 power, not 3.

Pluging in x(0) = 0 yields that:

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

[itex]x= ((1/3)t)^{1/3}[/itex] 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 [itex]x(0) = 0 \Rightarrow \dot{x}(0) = 0[/itex]

Pluging in [itex]\dot{x}[/itex] one gets:

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

and therefore the other solution is x = 0

Are my solutions correct?

x= 0 is definitely a solution!