Q1 answer: d y d x + 2 y = x 2 e − 2 x + 5 y = ∫ e ∫ P ( x ) d x Q ( x ) d x + C e ∫ P ( x ) d x e ∫ P ( x ) d x P ( x ) = 2 e ∫ 2 d x = e 2 x Q ( x ) = x 2 e − 2 x + 5 y = ∫ ( e 2 x ) ( x 2 e − 2 x + 5 ) d x + C e 2 x y = ∫ x 2 + 5 e 2 x d x + C e 2 x = x 3 3 + 5 e 2 x 2 + C e 2 x y = x 3 3 e 2 x + 5 2 + C e 2 x _ {\displaystyle {\begin{aligned}&{\text{Q1 answer:}}\\&{\frac {dy}{dx}}+2y={{x}^{2}}{{e}^{-2x}}+5\\&y={\frac {\int {{{e}^{\int {P\left(x\right)dx}}}Q\left(x\right)dx}+C}{{e}^{\int {P\left(x\right)dx}}}}\\&{{e}^{\int {P\left(x\right)dx}}}\\&P\left(x\right)=2\\&{{e}^{\int {2dx}}}={{e}^{2x}}\\&Q\left(x\right)={{x}^{2}}{{e}^{-2x}}+5\\&y={\frac {\int {\left({{e}^{2x}}\right)\left({{x}^{2}}{{e}^{-2x}}+5\right)}dx+C}{{e}^{2x}}}\\&y={\frac {\int {{{x}^{2}}+5{{e}^{2x}}}dx+C}{{e}^{2x}}}={\frac {{\frac {{x}^{3}}{3}}+{\frac {5{{e}^{2x}}}{2}}+C}{{e}^{2x}}}\\&{\underline {y={\frac {{x}^{3}}{3{{e}^{2x}}}}+{\frac {5}{2}}+{\frac {C}{{e}^{2x}}}}}\end{aligned}}}
