Let f(x) = x - 1 and g(x) = ex for x R. If dydx = (e-2 x g(f(f(x))) - y x ), y(0) = 0, then y(1) is
- 1-e3e4
- e-1e4
- 1-e2e4
- 2e-1e3
Answer
(2) e-1e4
Solution
f(f(x)) = (x - 1) - 1 = x - 2 g(f(f(x))) = ex-2
dydx + y x = ex - 2 - 2 x (linear differential equation)
Integrating factor = e 1 xdx = e2 x
ddx (y e2 x ) = ex - 2 - 2 x e2 x = ex-2
y e2 x = ex-2 + C
y(0) = 0 0 = e-2 + C C = -e-2
At x = 1: y(1) e2 = e-1 - e-2
y(1) = e-1 - e-2e2 = e - 1e4
Answer: (2)