Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Do you know separation of variables? –  Paul Oct 25 '12 at 13:20
1  
@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

2 Answers 2

up vote 3 down vote accepted

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

share|improve this answer
    
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

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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.