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First order DE

  • Thread starter fluidistic
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  • #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.
 

Answers and Replies

  • #2
tiny-tim
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Hi fluidistic! :smile:
1)[tex]\cdots 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2[/tex].
Nooo :redface:
 
  • #3
fluidistic
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Hi fluidistic! :smile:


Nooo :redface:
I got it thanks. Yeah a shame!
 
  • #5
fluidistic
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Last edited by a moderator:
  • #6
cronxeh
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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
fluidistic
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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
cronxeh
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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 lets 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
fluidistic
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Ok lets 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
fluidistic
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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
fluidistic
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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!
 

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