Question

y" =e^x+2x^2+4x+5
Y(0)=1
Y'(0)=4

Answer 1

$y''=e^x+2x^2+4x+5 \ y'=int (e^x+2x^2+4x+5)dx=e^x+ frac{2x^3}{3} +2x^2+5x+C_1 \ y=int (e^x+ frac{2x^3}{3} +2x^2+5x+C_1)dx=\=e^x+ frac{x^4}{6} +frac{2x^3}{3}+2.5x^2+C_1x+C_2 \\ y(0)=e^0+ frac{0^4}{6} +frac{2*0^3}{3}+2.5*0^2+C_1*0+C_2=1 \ 1+C_2=1 \C_2=0 \\ y'(0)=e^0+ frac{2*0^3}{3} +2*0^2+5*0+C_1=4 \1+C_1=4 \ C_1=3 \\y=e^x+ frac{x^4}{6} +frac{2x^3}{3}+2.5x^2+3x$