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Let$\frac{\text{dy}}{\text{d}x}=\frac{3y+1}{x^2}$

What is $y(x)$?

I tried anti-differentiation,but it seems does not work. Is there any tricks to solve the problem?

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    @Paul I don't know and I searched it in Wikipedia. It seems related to differential equation. Could you solve it?2012-10-25

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$\begin{align*}\frac{1}{3y+1}dy&=x^{-2}dx\\ \frac{1}{3}\int \frac{3}{3y+1} dy&=\int x^{-2}dx\\ \frac{1}{3}\ln|3y+1|&=-x^{-1}+C\\ 3y+1&=Ae^{-\frac{3}{x}}\text{, where }\ln A=3C\\ y&=\frac{1}{3}(Ae^{-\frac{3}{x}}-1)\\ y&=Be^{-\frac{3}{x}}-\frac{1}{3}\text{, where }B=\frac{1}{3}A \end{align*}$

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    @PENGTENG $\int \frac{1}{3y+1} dy=\frac{1}{3}\int \frac{3}{3y+1} dy$ The numerator of the fraction must be the derivative ($3$) of the denominator, in order to integrate it to $\ln$.2012-10-28
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$\frac{dy}{3y+1}=\frac{dx}{x^2}$

Integrate now both sides

$\frac{ln(3y+1)}{3}=\frac{-1}{x}+C$

$3y+1=De^{\frac{-3}{x}}$

$y=Ee^{\frac{-3}{x}}-\frac{1}{3}$