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

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

    $\displaystyle \frac{dx}{dt} = \sec^2 t = 1 + \tan^2 t = 1 + x^2$. 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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