Given y1(x)=e−x2\displaystyle {y}_{{1}}{\left({x}\right)}={e}^{{-{x}^{{2}}}} and y2(x)=xe−x2\displaystyle {y}_{{2}}{\left({x}\right)}={x}{e}^{{-{x}^{{2}}}} satisfy the corresponding homogeneous equation of

y′′+4xy′+(4x2+2)y=4e−x(x+2),t>0\displaystyle {y}{''}+{4}{x}{y}'+{\left({4}{x}^{{2}}+{2}\right)}{y}={4}{e}^{{-{x}{\left({x}+{2}\right)}}},\quad{t}>{0}

Then the general solution to the non-homogeneous equation can be written as y(x)=c1y1(x)+c2y2(x)+yp(x)\displaystyle {y}{\left({x}\right)}={c}_{{1}}{y}_{{1}}{\left({x}\right)}+{c}_{{2}}{y}_{{2}}{\left({x}\right)}+{y}_{{p}}{\left({x}\right)}.

Use variation of parameters to find yp(x)\displaystyle {y}_{{p}}{\left({x}\right)}.

yp(x)\displaystyle {y}_{{p}}{\left({x}\right)} =  

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