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Given that $2 x^{2}+\frac{3}{x}$ is a solution of the differential equation $y^{\prime \prime}-\frac{2 y}{x^{2}}=0$
Use the method of variation of parameters to find a particular solution of the equation $y^{\prime \prime}-\frac{2 y}{x^{2}}=3, x>0$

$y_{c}=c_{1} x^{2}+{c_{2}} \frac {1}{x}$

$y_{p}=A x^{2}+B \frac{1}{x}$

$A^{\prime} x^{2}+B^{\prime}\left(\frac{1}{x}\right)=0$

$* x$

$\rightarrow {A^\prime} x^{3}+B^{\prime}=0 \cdots \cdots (1)$

$2 {A^\prime} x-B^\prime \frac{1}{x^{2}}=3$

$* x^{2}$

$\longrightarrow 2 A^\prime x^{3}-B^\prime =3 x^{2} \cdots \cdots (2)$

$3 A^\prime x^{3}=3 x^{2}$

$\rightarrow A^{\prime}=\frac{1}{x}$ in $(1)$

$\frac{1}{x} x^{3}+B^{\prime}=0 \rightarrow B=-x^{2}$

$A=\int \frac{1}{x} d x=\ln |x|$

$B=\int-x^{2} d x=\frac{-x^{3}}{3}$

$y_{p}=x^{2} \ln |x|-\frac{x^{3}}{3} \cdot \frac{1}{x}$

$y_{p}=x^{2} \ln |x|-\frac{x^{2}}{3}$

$y_{g . s}=2 x^{2}+\frac{3}{x}+x^{2} \ln |x|-\frac{x^{2}}{3}$