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Use variation of parameters to find the particular solution $$y(x)$$ of the equation:
$$y^{\prime \prime}-2 y^{\prime}+y=\frac{e^{x}}{x}$$
Such that $$y(1)=0, y^{\prime}(1)=0$$

$$f(m)=0 \quad \longrightarrow f(m)=m^{2}-2 m+1=0$$


$$(m-1)^{2}=0 \quad m-$$ values $$=+1,+1 \quad y_{c}=c_{1} e^{+x}+c_{2} x e^{+ x} $$

$$y p=A e^{x}+B x e^{x} $$

$$A^\prime e^{x}+B^\prime x e^{x}=0 \longrightarrow A^{\prime}=-B^{\prime} x$$

$$A^\prime e^{x}+B^{\prime} e^{x}+B^{\prime} x e^{x}=\frac{e^{x}}{x} $$

$$-B^{\prime} x e^{x}+B^{\prime} e^{x}+B^{\prime} x e^{x}=\frac{e^{x}}{x} $$

$$\longrightarrow-B^{\prime} x+B^{\prime}+B^{\prime} x=\frac{1}{x} $$

$$B^\prime=\frac{1}{x} $$

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

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

$$A=\int-1 d x=-x$$

$$y_{p}=-x e^{x}+\ln |x| x e^{x}$$

$$y_{g . s}=y_{c}+y_{p}=-x e^{x}+\ln |x| e^{x}+c_{1} e^{x}+c_{2} x e^{x} $$

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

$$y_{g . s}=c_{1} e^{x}+c_{2} x e^{x}-x e^{x}+x e^{x} \ln |x|$$

$$y(1)=0 \rightarrow y=\frac{c_{1} e+c_{2} e-e+0}{e} $$

$$\rightarrow c_{1}+c_{2}-1=0 \rightarrow c_{1}+c_{2}=1 \cdots (1)$$

$$y^{\prime}(1)=0 \rightarrow y^{\prime}=c_{1} e^{x}+c_{2} e^{x}+ c_{2} x e^{x}-e^{x}-x e^{x}+e^{x} \ln |x|+e^{x} \ln |x|+\frac{1}{x} x e^{x} $$

$$y^{\prime}(1)=c_{1} e+c_{2} e+c_ 2 e-e-e+e+e \ln |1|+e(1) \ln | 1 |+e$$

$$=c_{1} e+2 c_{2} e-e=0$$

$$\div e$$

$$2 c_{2}+c_{1}-1=0 \rightarrow 2 c_{2}+c_{1}=1 \cdots (2)$$

$$C_{1}+C_{2}=1 \rightarrow C_{1}=1-C_{2} \text {            } in(2)$$

$$2 C_{2}+1-C_{2}=1$$


$$y_{g . s}=e^{x}+-x e^{x}+x e^{x} \ln |x|$$

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