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Math Help - What am I doing wrong, differential equation

  1. #1
    MHF Contributor arbolis's Avatar
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    What am I doing wrong, differential equation

    We're looking for x(t). The data are : x'=f(t,x), x(0)=0.
    We have f(t,x)=1+x^2.
    I rewrite it as \frac{dx}{dt}=1+x^2. Which is equivalent to 1=\frac{dx}{(1+x^2)dt}. Equivalent to \int dt=\int \frac{dx}{1+x^2}. Which means that  t=arctan(x). As we were looking for x(t), from what we got it's easy to see that x(t)=tan(t). x(0)=0 is satisfied.
    Now looking at the beginning, we had that x'(t)=1+x^2. What I'm missing to understand is how my solution works here? If I derivate tan(t) I won't get the 1+x^2. I'm confused.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by arbolis View Post
    We're looking for x(t). The data are : x'=f(t,x), x(0)=0.
    We have f(t,x)=1+x^2.
    I rewrite it as \frac{dx}{dt}=1+x^2. Which is equivalent to 1=\frac{dx}{(1+x^2)dt}. Equivalent to \int dt=\int \frac{dx}{1+x^2}. Which means that  t=arctan(x). As we were looking for x(t), from what we got it's easy to see that x(t)=tan(t). x(0)=0 is satisfied.
    Now looking at the beginning, we had that x'(t)=1+x^2. What I'm missing to understand is how my solution works here? If I derivate tan(t) I won't get the 1+x^2. I'm confused.
     t = \arctan(x) + C. Impose the boundary condition: C = 0.

    So  t = \arctan(x) \Rightarrow x = \tan (t). So all is well.

    \frac{dx}{dt} = \sec^2 t = 1 + \tan^2 t = 1 + x^2. So there's no problem - things work just fine.
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  3. #3
    MHF Contributor arbolis's Avatar
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    . So there's no problem - things work just fine.
    That may looks innocent but now I understand what I wasn't understanding! Great.
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