Find the Solution for x' = t + x/(1+x^2) with Initial Condition x(0) = 0

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In summary: Hello Hall. How are you? May I say I'm not satisfied by this outcome? I know what you're thinkin': "why do I even bother; he's a pain in the . . .". Tell you what, suppose an asteriod was heading here and we had to find some analytical expression for this ODE in order to successfully deflect it. What progress could the combined intellect of the world muster to do so? I bet a whole dollar something could be done. :smile:
  • #1
twoflower
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Hi all,

I found this problem on web and though we haven't encountered this kind of problem in class I wonder how could it be done. Here it is:


Guess solution of the problem


[tex]
x' = t + \frac{x}{1+x^2},\mbox{ x(0) = 0}
[/tex]

ie. find functions [itex]\omega(t) \leq x(t) \leq \phi(t)[/itex] for each [itex]t[/itex] from domain of solution.

Well, I have no clue. I have the results here if anyone tries...

Thank you for hints!
 
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  • #2
HINT: What are the largest and smallest possible values of the real function

[tex]\frac {x}{1+x^2}[/tex]
?
 
  • #3
Tide said:
HINT: What are the largest and smallest possible values of the real function
[tex]\frac {x}{1+x^2}[/tex]
?

[tex]
\frac{1}{2}
[/tex]

and

[tex]
-\frac{1}{2}
[/tex] ?
 
  • #4
Yes, and if you replace the "x" in the right hand side of the equation by those, can you then solve for x(t)?
 
  • #5
HallsofIvy said:
Yes, and if you replace the "x" in the right hand side of the equation by those, can you then solve for x(t)?

Thank you HallsofIvy, so I wrote the bounds

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

and

[tex]
x \geq \frac{t^2}{2} - \frac{t}{2}
[/tex]

It's ok, isn't it? Anyway, the official results say something slightly different...
 
  • #6
twoflower said:
It's ok, isn't it? Anyway, the official results say something slightly different...


Ok, I have it. In the results, there is

[tex]
\frac{t^2}{2} - t \leq x(t) \leq \frac{t^2}{2} + t
[/tex]

which is just result of rougher bounds.
 
  • #7
So can someone suggess a method for solving the original ODE analytically? I can't.
 
  • #8
Non-linear differential equations, such as this one, tend NOT to have solutions that can be found exactly.
 
  • #9
HallsofIvy said:
Non-linear differential equations, such as this one, tend NOT to have solutions that can be found exactly.

Hello Hall. How are you? May I say I'm not satisfied by this outcome? I know what you're thinkin': "why do I even bother; he's a pain in the . . .". Tell you what, suppose an asteriod was heading here and we had to find some analytical expression for this ODE in order to successfully deflect it. What progress could the combined intellect of the world muster to do so? I bet a whole dollar something could be done. :smile:
 

FAQ: Find the Solution for x' = t + x/(1+x^2) with Initial Condition x(0) = 0

What is the method used to solve this differential equation?

The method used to solve this differential equation is separation of variables.

How do you find the general solution for x' = t + x/(1+x^2)?

To find the general solution, we first rearrange the equation to separate the variables on opposite sides. This gives us dx/(1+x^2) = dt. We can then integrate both sides with respect to their corresponding variables. This will give us ln(1+x^2) = t^2/2 + C, where C is the constant of integration. We can then solve for x by taking the exponential of both sides, giving us x = sqrt(e^(t^2/2 + C) - 1).

How do you find the particular solution for x' = t + x/(1+x^2) with initial condition x(0) = 0?

To find the particular solution, we substitute the initial condition into the general solution. In this case, when t = 0 and x = 0, we get C = 0. This gives us the particular solution x = sqrt(e^(t^2/2) - 1).

Can this differential equation be solved analytically?

Yes, this differential equation can be solved analytically using the method of separation of variables.

What is the significance of the initial condition x(0) = 0 in this problem?

The initial condition represents the starting point for the solution. In this problem, it means that at time t = 0, the value of x is 0. This condition allows us to find the particular solution and determine the behavior of x at different time intervals.

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