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