Question

find the particular solution that satisfies the differential equation and the initial condition

f'(x) =2x+4, f(-2) =3

Answer 1

f'(x) =2x+4

Take integration both sides

∫f'(x)dx =∫(2x+4)dx

f(x) =(2(x^2/2))+4x+C

f(x) =x^2+4x+C

f(-2) =3

3=(-2)^2+4(-2)+C

C =3-4+8 =7

f(x) =x^2+4x+7