Question

If f(x) is a differentiable function then the solution of dy + (yf'(x) – f(x) f'(x)) dx = 0 is

• A. y = (f(x) – 1) + ce^–f(x)
• B. yf(x) = (f(x))² + c
• C. ye^f(x) = f(x) e^f(x) + c
• D. (y – f(x)) = f(x) e^–f(x)

Answer 1

Answer: (A)

y = (f(x) – 1) + ce^–f(x)

Answer 2

$\frac{dy}{dx}$ = – [y – f(x)] f '(x)

Put y – f(x) = z $\frac{dy}{dx}$ – f'(x) = $\frac{dz}{dx}$

f '(x) + $\frac{dz}{dx}$ = – zf '(x)

$\frac{dz}{1 + z}$ = – f '(x) dx

log (1 + z) = – f(x) + k 1 + z = e^{– f(x) + k}

1 + y – f(x) = ce^–f(x)

y = f(x) – 1 + ce^–f(x)