Q1 answer:dydx+2y=x2e−2x+5y=∫e∫P(x)dxQ(x)dx+Ce∫P(x)dxe∫P(x)dxP(x)=2e∫2dx=e2xQ(x)=x2e−2x+5y=∫(e2x)(x2e−2x+5)dx+Ce2xy=∫x2+5e2xdx+Ce2x=x33+5e2x2+Ce2xy=x33e2x+52+Ce2x_{\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}}}