Question

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

Answer 1

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