y'=x-y, y(0)=1. f(x)=x-1+2e^-x

Question

use differentiation and substitution to show that the function y=f(x) is the exact solution of the initial value problem.

Answer 1

dy/dx = x - y

y(x) = x - 1 + 2e^-x

x - y(x) = x - x + 1 - 2e^-x

x - y(x) = 1 - 2e^-x

y'(x) = dy/dx = 1 + 2e^-x = x - y(x)