I'm trying to solve
$$x^2y''+2xy'-12y =\sqrt{x}$$
I have found and verified the solution,
$$y(x) =\frac{c_2}{x^4}+c_1 x^3 -\frac{4}{45}\sqrt{x}$$
via wolfram, however I can't get there manually...
I have $y_1 = x^3, y_2=x^{-4}, W[y_1,y_2] = -7x^{-2}$
and I'm using the formula $y_p = y_1v_1 +y_2v_2$
where
$${v_1}' =\frac{-x^{-4}x^{1/2}}{-7x^{-2}} = \frac{1}{7}x^{-3/2}$$
$$v_1 = -\frac{2}{7}x^{-1/2}$$
and
$${v_2}' = \frac {x^3x^{1/2}}{-7x^{-2}} = \frac{-2}{91}x^{11/2}$$
$$v_2 = -\frac{2}{91}x^{13/2}$$
so my particular solution came out to be:
$$y_p = \frac{-2}{7}x^{5/2} + \frac{-2}{91}x^{5/2}$$
(which is different from $x^{1/2}$ from above)
I know it's probably a simple mistake somewhere but I really can't find it.. :/
You have to divide the differential equation by the leading therm's coefficient, in this case $x^2$ so \begin{equation} \frac{d^2y(x)}{dx^2}+\frac{2\frac{d^2y(x)}{dx^2}}{x^2}=\frac{1}{x^{3/2}} \end{equation} So you can compute \begin{equation} v_1(x)=\int\frac{x^{7/2}}{7}dx=-\frac{2x^{9/2}}{63} \end{equation} and \begin{equation} v_2(x)=\int\frac{1}{7x^{7/2}}dx=-\frac{2}{35x^{5/2}} \end{equation} So, the particular solution is $v_1y_1+v_2y_2=-\frac{2 \sqrt{x}}{63}-\frac{2\sqrt{x}}{35}$ If you simplify you will get to $y_p(x)=-\frac{4\sqrt{x}}{45}$
***I think I took the opposite $y_1$ and $y_2$ you've computed, but result still must be the same