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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 $$(1)$$

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

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

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

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

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

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

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