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 $y_p = y_1v_1 +y_2v_2$


$${v_1}' =\frac{-x^{-4}x^{1/2}}{-7x^{-2}} = \frac{1}{7}x^{-3/2}$$

$$v_1 = -\frac{2}{7}x^{-1/2}$$


$${v_2}' = \frac {x^3x^{1/2}}{-7x^{-2}} = \frac{-2}{91}x^{11/2}$$

$$v_2 = -\frac{2}{91}x^{13/2}$$

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

  • $\begingroup$ Where? the formula is $v_2(x)=\int\frac{f(x)y_1(x)}{W(x)}dx$, here $f(x)=\frac{1}{x^{3/2}}$ and $y_1(x)=x^{-4}$, said i took the opposite y's $\endgroup$ – Karen May 6 '18 at 22:56
  • $\begingroup$ also @sksumii13 you must check your answer since the last therm is incorrect $\endgroup$ – Karen May 6 '18 at 23:03
  • $\begingroup$ wolfram answer corrected 😅 $\endgroup$ – sksumii13 May 7 '18 at 2:00
  • $\begingroup$ fine, hope my answer helped you $\endgroup$ – Karen May 7 '18 at 4:29

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