MHB How to approach this 2nd order linear ODE?

[ineedhelpnow]

$y''+\frac{1}{x}y'=\frac{2}{x^2}-4$

Hey. This is probably really simple but I'm stuck :p how do I approach this?

 

[MarkFL]

I would let:

$$u=\d{y}{x}$$

So that the ODE becomes:

$$u'+\frac{1}{x}u=\frac{2}{x^2}-4$$

Now, you should recognize this form...something about an integrating factor?

 

[MarkFL]

Since no feedback from the OP has been given for a couple of days, I am going to git-r-done. If we multiply through by $x$, we obtain:

$$xu'+u=\frac{2}{x}-4x$$

Something wonderful has transpired here, for behold, the left side of the ODE may now be written as:

$$\frac{d}{dx}\left(xu\right)=\frac{2}{x}-4x$$

Now, upon integrating through with respect to $x$, we obtain:

$$ux=2\ln(x)-2x^2+c_1$$

Dividing through by $x$, we find:

$$u=\frac{2}{x}\ln|x|-2x+\frac{c_1}{x}$$

Back-substitute for $u$:

$$\d{y}{x}=\frac{2}{x}\ln|x|-2x+\frac{c_1}{x}$$

Integrating again with respect to $x$, we now obtain the general solution to the original ODE:

$$y(x)=\ln^2|x|-x^2+c_1\ln|x|+c_2$$