method to solve a Second order of differential equation

Question

$$y''(x)+\left(\dfrac 2 x +\dfrac x 2\right)y'(x)+ \dfrac 3 2 y(x)=0$$

Can anyone help me by telling me a method to solve this kind of differential equations ? I tried Cauchy Euler method, but it did not work.

Answer 1

The term $\frac 2x+\frac x2=\frac{x^2+4}{2x}$ gave me the idea to define $y=e^{-\frac{x^2}{4}} z$ which makes the equation $$e^{-\frac{x^2}{4}}\frac{ \left(4-x^2\right) z'(x)+2 x z''(x)}{2 x}=0$$ So, consider $$\left(4-x^2\right) z'+2 x z''=0$$ Now, reduce the order $u=z'$ to get $$\left(4-x^2\right) u+2 x u'=0$$ which is separable and leads to $$u=z'=\frac{c_1}{x^2} e^{\frac{x^2}{4}}\implies z=c_1 \left(\frac{1}{2} \sqrt{\pi } \text{erfi}\left(\frac{x}{2}\right)-\frac{e^{\frac{x^2}{4}}}{x}\right)+c_2$$

Answer 2

$$y''(x)+(\dfrac 2x +\dfrac x2)y'(x)+ (\dfrac32) y(x)=0$$ Multiply by $2x^2$ $$(2x^2y')'+(x^3y)'=0$$ Integrate $$2x^2y'+x^3y=K$$ $$y'+\frac x2y=\frac K {2x^2}$$ $$y'+\frac x2y=\frac C{x^2}$$ You can try to solve the first ode ( It may not be expressed with elementary functions) $$(ye^{x^2/4})'=e^{x^2/4}\frac C{x^2}$$ $$y=C_1e^{-x^2/4} \int \frac {e^{x^2/4}}{x^2}dx+C_2e^{-x^2/4}$$