# Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method

Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method

By integrating factor method -

I put it into the form of $\frac{dy}{dx} + \frac{4x}{x^2 + 1} \cdot y = \frac{x}{x^2 + 1}$

My integrating factor is $I = (x^2 + 1)^2$

By formula, $y (x^2 +1)^2 = \int{ (x^3 + x)} dx$

And thus my general solution is $y = \frac{x^4 + 2x^2 + C}{4(x^2 + 1)^2}$

By separable method is where I got problems with...

I put it into the form of $\int{ (\frac{x}{x^2 + 1} )} dx = \int{ (\frac{1}{(1-4y)}} dy$

Getting $\frac{1}{2} \ln (x^2 + 1) + C = \frac{1}{4} \ln (1-4y)$

$2 \ln (x^2 +1) + C = \ln (1-4y)$

$1-4y = e^{\ln (x^2 + 1)^2} + e^{C} = (x^2 +1)^2 + C$

$y = \frac{ (x^2+1)^2 + C - 1}{4}$

Why is my general solution from separable method very different from the integrating factor method ? I suspect that I have went wrong in the separable method but cannot identify it ...

$$\displaystyle \int \dfrac{1}{1-4y}~dy=\int\dfrac{x}{x^{2}+1}~dx$$
$\dfrac{-1}{4}\log(1-4y)=\dfrac{1}{2}\log(x^{2}+1)$ You have missed minus sign.