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• Notes

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 $y(1)=0, y^{\prime}(1)=0$

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

$(m-1)(m-1)=0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\div e$

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

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

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

$c_{1}=1-0=1$

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