I have trouble solving the following ODE: $$y''(x)+\frac{1}{2}xy'(x)-\frac{3}{2}y=0.$$ Could someone help me with this question? Thanks!
2026-04-24 14:30:40.1777041040
Solve second order linear homogeneous ODE with variable coefficients
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For a given differential equation ${y}''\left( x \right)+P\left( x \right){y}'\left( x \right)+Q\left( x \right)y\left( x \right)=0$
If a solution ${{y}_{1}}\left( x \right)$ known, then ${{y}_{2}}\left( x \right)$ is given by
${{y}_{2}}\left( x \right)={{y}_{1}}\left( x \right)\int{\frac{{{e}^{\int{-P\left( x \right)dx}}}}{{{y}_{1}}{{\left( x \right)}^{2}}}}dx$
So as you suggested ${{y}_{1}}\left( x \right)=\left( {{x}^{3}}+6x \right)$ is a solution, hence
${{y}_{2}}\left( x \right)=\left( {{x}^{3}}+6x \right)\int{\frac{{{e}^{-\frac{{{x}^{2}}}{4}}}}{{{\left( {{x}^{3}}+6x \right)}^{2}}}}dx={{\text{e}}^{-\frac{{{x}^{2}}}{4}}}\left( -\frac{x}{72\left( {{x}^{2}}+6 \right)}-\frac{1}{36x} \right)-\frac{\sqrt{\pi }}{48}\,erf\left( \frac{x}{2} \right)$
Where $erf\left( z \right)=\frac{2}{\sqrt{\pi }}\underset{0}{\overset{z}{\mathop \int }}\,{{e}^{-{{t}^{2}}}}dt$ the error function.