Solving ODEs Passing Through Points: x'=x^{\frac{1}{2}}

In summary, the conversation discusses finding solutions for the differential equation x'=x^(1/2) that pass through given points, as well as determining the initial conditions and solving for the constant in the solution expression. The conversation also touches on the use of Wolfram Alpha and the confusion surrounding the notation of x_0 and t_0.
  • #1
fluidistic
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Homework Statement


1)Find the solution of [tex]x'=x^{\frac{1}{2}}[/tex] that passes through the point [tex](t_0, x_0)[/tex] where [tex]x_0>0[/tex].
2)Find all the solutions of this equation that pass through the point [tex](t_0,0)[/tex].

Homework Equations


Direct integration.


The Attempt at a Solution


1)[tex]\frac{dx}{dt}=x^{\frac{1}{2}} \Rightarrow \frac{dx}{x^{\frac{1}{2}}}=dt \Rightarrow \int \frac{dx}{x^{\frac{1}{2}}}=t+C\Rightarrow 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2[/tex].

I determined C thanks to the initial condition and the equation became [tex]x=\frac {t^2}{4}+t \left ( \frac{2x_0 ^{\frac{1}{2}}-t_0}{2}} \right ) + \frac{(2x_0 ^{\frac{1}{2}}-t_0)^2}{4}}[/tex].
2) Replacing [tex]x_0[/tex] by [tex]0[/tex] in the above equation yields [tex]x= \left ( \frac{t}{2}-\frac{t_0}{2} \right ) ^2[/tex].
Unfortunately I replaced this solution into the original equation and the equality isn't true. So I made an error. I also replaced the first solution I got (the one with C's) into the equation and it didn't work. Hence I made an error quite early. I don't know where though.
 
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  • #2
Hi fluidistic! :smile:
fluidistic said:
1)[tex]\cdots 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2[/tex].

Nooo :redface:
 
  • #3
tiny-tim said:
Hi fluidistic! :smile:


Nooo :redface:

I got it thanks. Yeah a shame!
 
  • #6
fluidistic said:
I don't hate those. I knew I was wrong as I stated in my first post. I didn't know where though. Wolfram alpha would have showed that I was wrong early, but this, I already knew. Thanks anyway.

Please post your final solution. There are some mistakes that make me go :eek:
 
  • #7
cronxeh said:
Please post your final solution. There are some mistakes that make me go :eek:
Ok. I have at least one doubt.
Here goes my attempt: [tex]x=\frac{t^2+2tC+C^2}{4}[/tex]. It satisfies the DE, so I'm sure I'm right until here.
Now to answer 1): [tex]x_0=x(t_0)=\frac{t_0^2+2t_0C+C^2}{4}[/tex].
Thus [tex]C^2+2t_0C+(t_0^2-4x_0)=0[/tex]. Solving for C, I got [tex]C=-t_0 \pm 2 x_0^{\frac{1}{2}}[/tex].
Hence [tex]x= \frac{t^2+2t (2x_0^ {\frac{1}{2}} -t_0)+(2x_0^{\frac{1}{2}}-t_0 )^2}{4}[/tex].
Notice that I took [tex]C=-t_0+2 x_0^{\frac{1}{2}}[/tex] but I'm not sure why I didn't take [tex]C=-t_0-2 x_0^{\frac{1}{2}}[/tex]. This is my doubt.
To answer 2), I just plugged [tex]x_0=0[/tex] in the above equation to reach oh... yeah the same wrong result I had.
I'm really clueless where I made an error. I revised the quadratic in C and I don't see any error. The first equation is right... so, maybe my method isn't. I'd love to know where are my mistakes.
 
  • #8
fluidistic said:
Ok. I have at least one doubt.
Here goes my attempt: [tex]x=\frac{t^2+2tC+C^2}{4}[/tex]. It satisfies the DE, so I'm sure I'm right until here.
Now to answer 1): [tex]x_0=x(t_0)=\frac{t_0^2+2t_0C+C^2}{4}[/tex].
Thus [tex]C^2+2t_0C+(t_0^2-4x_0)=0[/tex]. Solving for C, I got [tex]C=-t_0 \pm 2 x_0^{\frac{1}{2}}[/tex].
Hence [tex]x= \frac{t^2+2t (2x_0^ {\frac{1}{2}} -t_0)+(2x_0^{\frac{1}{2}}-t_0 )^2}{4}[/tex].
Notice that I took [tex]C=-t_0+2 x_0^{\frac{1}{2}}[/tex] but I'm not sure why I didn't take [tex]C=-t_0-2 x_0^{\frac{1}{2}}[/tex]. This is my doubt.
To answer 2), I just plugged [tex]x_0=0[/tex] in the above equation to reach oh... yeah the same wrong result I had.
I'm really clueless where I made an error. I revised the quadratic in C and I don't see any error. The first equation is right... so, maybe my method isn't. I'd love to know where are my mistakes.

Ok let's rewrite it as:

x(t) = 1/4 (2*C*t + C^2 + t^2)

x_0 means x(0) means t=0

x(0) = 1/4*C^2 C = sqrt(4*x0) = 2*sqrt(x0)

So for problem #1 you are asked to find x(t) = 1/4* (2*2*sqrt(x0)*t + 4*x0 + t^2)

and x(t) = 1/4*(4*sqrt(x0)*t + 4*x0 + t^2)

for problem #2 you are asked to find solutions that pass through (t0,x0=0)

x(t) = 1/4*(4*sqrt(0)+4*0+t^2) = 1/4 * t^2 = (t^2)/4
 
  • #9
cronxeh said:
Ok let's rewrite it as:

x(t) = 1/4 (2*C*t + C^2 + t^2)

x_0 means x(0) means t=0

x(0) = 1/4*C^2 C = sqrt(4*x0) = 2*sqrt(x0)

So for problem #1 you are asked to find x(t) = 1/4* (2*2*sqrt(x0)*t + 4*x0 + t^2)

and x(t) = 1/4*(4*sqrt(x0)*t + 4*x0 + t^2)

for problem #2 you are asked to find solutions that pass through (t0,x0=0)

x(t) = 1/4*(4*sqrt(0)+4*0+t^2) = 1/4 * t^2 = (t^2)/4
Thanks a lot for the answer, I'm going back to it as soon as I can, i.e. in about 2 days... I have to do a trip of more than 800 km (11 hours) due to a passport thing.
 
  • #10
I'm back!
Ok, it seems much easier than I thought. But I have some questions, why did you chose [tex]C=2x_0 ^{\frac{1}{2}}[/tex] and not [tex]C=-2x_0 ^{\frac{1}{2}}[/tex]?
My other questions are that although I misunderstood what was meant by [tex]x_0[/tex] (I thought it was x(t_0) instead of x(0) ), why won't my final expression of x(t) work? I should have found a particular solution to the equation... or not?
Also, what's the deal with this [tex]t_0[/tex]? It doesn't even appear a single time anywhere but in the question.
 
  • #11
Small bump... can someone answer my last 3 questions? I'm self studying DE's, I really need to understand the topic and my 3 last questions are puzzling me.
Thanks in advance for any help!
 

1. What is an ODE?

An ODE (ordinary differential equation) is an equation that relates a function to its derivatives. It is commonly used to model dynamic systems in physics, engineering, and other scientific fields.

2. How do I solve an ODE?

There are various methods for solving ODEs, such as separation of variables, substitution, and using an integrating factor. The specific method used depends on the form of the ODE and its initial conditions.

3. What is the function x'=x^{\frac{1}{2}}?

This function is a first-order ODE, where the derivative of x (denoted by x') is equal to the square root of x. It represents a system where the rate of change of x is proportional to its square root.

4. What does it mean to solve an ODE passing through specific points?

Solving an ODE passing through specific points means finding the unique solution to the ODE that satisfies the given initial conditions. In other words, the solution curve must pass through the given points.

5. How can I use the function x'=x^{\frac{1}{2}} to solve a problem?

This function can be used to model systems where the rate of change is proportional to the square root of the variable. It can also be used to solve certain types of real-world problems, such as population growth or radioactive decay.

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