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, so g(f(f(x))) = ex-2 and the equation is the linear differential equation
dydx + y x = ex - 2 - 2 x
Its integrating factor is e 1 xdx = e2 x, so
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, so C = -e-2.
At x = 1, y(1) e2 = e-1 - e-2
y(1) = e-1 - e-2e2 = e - 1e4
Answer: (2)