Simple 1st order ODE. Need insight

[Samuelb88]

Homework Statement

Solve the 1st order linear initial value problem.

$$(1+x^2)\frac{dy}{dx}\right) + 4xy = x$$ , $$y(1)=\frac{1}{4}\right)$$

The Attempt at a Solution

Why can I get the answer solving the equation one way (below), but not they other?

Method 1:
$$(1+x^2)\frac{dy}{dx}\right) = -x(4y-1)$$

$$\frac{dy}{dx}\right) * \frac{1}{4y-1}\right) = -\frac{x}{1+x^2}\right)$$

$$\frac{1}{4}\right) \int^{}_{} \frac{1}{4y-1}\right) d(4y-1) = -\frac{1}{2}\right) \int^{}_{} \frac{1}{1+x^2}\right d(1+x^2)$$

(multiply by 4)

$$ln|4y-1| = -2ln|1+x^2| + C$$

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

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

Satisfying y(0)=1/4, C=0

Therefore: y = 1/4Method 2:

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

$$\frac{dy}{dx}\right) * \frac{1}{4y-1}\right) = -\frac{x}{1+x^2}\right)$$

$$\frac{1}{4}\right) \int^{y(x)}_{1/4} \frac{1}{4y-1}\right) d(4y-1) = -\frac{1}{2}\right) \int^{x}_{1} \frac{1}{1+x^2}\right d(1+x^2)$$

$$( ln|4y-1| - ln|0| ) = -2(ln|1+x^2| - ln|2|)$$

Here's where I'm confused. should I ignore ln0? or should I solve with infinity in my equation.

 

[clamtrox]

Remember, don't divide by zero :)