# Let$\frac{\text{dy}}{\text{d}x}=\frac{3y+1}{x^2}$ ,What is $y(x)$?

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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Do you know separation of variables? – Paul Oct 25 '12 at 13:20
@Paul I don't know and I searched it in Wikipedia. It seems related to differential equation. Could you solve it? – John Hass Oct 25 '12 at 13:22

\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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Could you please briefly explain why you multiply $\frac{1}{3}$ on the left hand expression in the second row? Thanks : ) – John Hass Oct 28 '12 at 16:20
@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 $\ln$. – Alex Oct 28 '12 at 16:58

$$\frac{dy}{3y+1}=\frac{dx}{x^2}$$

Integrate now both sides

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

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

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

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