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:

\(\displaystyle u=\d{y}{x}\)

So that the ODE becomes:

\(\displaystyle u'+\frac{1}{x}u=\frac{2}{x^2}-4\)

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:

\(\displaystyle xu'+u=\frac{2}{x}-4x\)

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

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

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

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

Dividing through by $x$, we find:

\(\displaystyle u=\frac{2}{x}\ln|x|-2x+\frac{c_1}{x}\)

Back-substitute for $u$:

\(\displaystyle \d{y}{x}=\frac{2}{x}\ln|x|-2x+\frac{c_1}{x}\)

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

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