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Is there a closed form solution for this first order ODE of $x(t)$ where $a(t)$ is as smooth as needed? $$x'(t)=\frac{a(t)}{x(t)}+b$$ Here, $b$ is a constant. But what if $b$ is also a function of $t$? But I do not want a series expansion formula with recursion equations resulting from the Frobenius method.

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    $\begingroup$ It obviously depends on what functions are $a$, $x$ and $b$ $\endgroup$ – luka5z Sep 5 '14 at 19:34
  • $\begingroup$ @luka5z: Sure, but I am wondering if $x(t)$ depends just on something simple, say $\int a(t)$ or $a'(t)$. $\endgroup$ – Hans Sep 5 '14 at 19:36
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This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $x(t)=\dfrac{1}{y(t)}$ ,

Then $x'(t)=-\dfrac{y'(t)}{(y(t))^2}$

$\therefore-\dfrac{y'(t)}{(y(t))^2}=a(t)y(t)+b(t)$

$y'(t)=-a(t)(y(t))^3-b(t)(y(t))^2$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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  • $\begingroup$ I did try the $y$ transformation but I did not know Abel equation. Thank you, doraemonpaul. $\endgroup$ – Hans Sep 5 '14 at 23:10

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