за решение

[Гост]

y"-2y`+y=[tex]\frac{e^{x}}{\sqrt{4-x^{2}}}[/tex]

[Добромир Глухаров]

$y''-2y'+y=\frac{e^x}{\sqrt{1-x^2}}$

$y''-2y'+y=0$

$y=e^{r.x}$

$(r^2-2r+1)e^{rx}=0$

$(r-1)^2=0$

$r_{1,2}=1$

$y=Y+\eta$; $Y=C_1.e^x+C_2.x.e^x$

$\eta=u_1(x)e^x+xu_2(x)e^x$

$\begin{array}{|l}e^xu_1'(x)+xe^xu_2'(x)=0\\e^xu_1'(x)+(e^x+xe^x)u_2'(x)=\frac{e^x}{\sqrt{1-x^2}}\end{array}$

$\begin{array}{|l}u_1'=-xu_2'\\u_1'+(1+x)u_2'=\frac{1}{\sqrt{1-x^2}}\end{array}$

$-xu_2'+(1+x)u_2'=\frac{1}{\sqrt{1-x^2}}$

$u_2'=\frac{1}{\sqrt{1-x^2}}\Rightarrow u_2=arcsinx+C$

$u_1'=-\frac{x}{\sqrt{1-x^2}}\Rightarrow u_1=\frac{1}{2}\int(1-x^2)^{-\frac{1}{2}}d(1-x^2)=\sqrt{1-x^2}+C$

$y=(C_1+xC_2)e^x+\sqrt{1-x^2}e^x+x.arcsinx.e^x$