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I have this equation that I have to solve, i try the most common methods but I do not get anywhere. I do not know if I'm doing something wrong or I need some more complex technique.

$y'=1-(1+y^2)^{3/2}$

can someone tell me how to solve it? regards

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  • $\begingroup$ The equation is separable. $\endgroup$
    – amWhy
    Aug 26, 2018 at 15:03
  • $\begingroup$ This is a separable equation, so there is a solution in the form $\int f(y) dy = x+C$. It may be difficult to explicitly solve this for $y$. $\endgroup$
    – Kusma
    Aug 26, 2018 at 15:03

2 Answers 2

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By separation, you can see that the solution $y(x)$ is given implicitly by

$$x = C + \int \frac{\text dy}{1-(1+y^2)^{\frac{3}{2}}} $$

Unfortunately, like most nonlinear equations, there's not much more you can do with it in terms of arriving at an explicit solution. You can evaluate the integral, but it will be extremely messy, and will not really get you closer to an explicit form.

You can, however, still find out plenty of interesting information about the behavior of $y$ without having an explicit form. For example, it's clear that $y'=0$ when $x=0$ and $y'<0$ otherwise. This tells you that, when $y(0)\geq0$, $\lim\limits_{x\to\infty} y(x) = 0$, and otherwise $y$ is a decreasing negative function of $x$ with no lower bound. This should allow you to sketch the graph of $y(x)$ for different values of $y(0)$ (or equivalently, different values of $C$ in the implicit solution).

And just a general tip, if an equation is separable but the integral is something crazy or undoable, chances are any other technique is going to come up with an equally crazy looking or unfindable solution. The only possible exceptions would be if that crazy integral is some constant difference from a reasonable function, which is highly unlikely.

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$$y'=1-(1+y^2)^{3/2}$$

We have $$\int \frac {dy}{1-(1+y^2)^{3/2}} =\int dx$$

With $y=\tan t$ we get

$$ \int \frac {dy}{1-(1+y^2)^{3/2}} = \int \frac {\sec ^2 t dt }{1+\sec ^3 t } = \int \frac {\cos t dt }{1+\cos ^3 t } $$ You can take over from here.

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  • $\begingroup$ in less than two lines !! I was impressed after trying to solve it and read the previous comment thank you very much $\endgroup$
    – Mathinho
    Aug 26, 2018 at 18:11

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