If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$?

Question

If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$

(a) $1/2e^{-2x+2}$

(b) $e^{-x-1}$

(c) $e^{1-x}$

(d) $e^{-x}$

(e) $-e^{x}$

If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.

Answer 1

Hint: All you need to do in this case is to evaluate $f(1)$: $f_A(1), f_B(1), f_C(1), f_D(1), f_E(1)$.
Show your calculations in each case.

Doing that rules out all possibilities except $(C)$, irregardless of the first condition, though you want to evaluate $f'_C(x)$ to show that $f_C = f'_C$.

Answer 2

If $f'(x) = -f(x)$, then $0 = f'(x)+f(x) = e^x(f'(x)+f(x)) = (e^x f(x))' $ so $e^x f(x)$ is constant.

Set $e^x f(x) = c$ and set $x = 1$. Then, since $f(1) = 1$, $c = e f(1) = e$ so $f(x) = e e^{-x} = e^{1-x}$.

An alternative solution is to write $f'(x)/f(x) = -1$ so $(\ln(f(x))'=-1$ so $\ln f(x) = c-x$ or $f(x) = e^{c-x}$ for some constant $c$.

This has the problem that $f(x) = 0$ causes problems, and the first solution does not have this problem.

Answer 3

This is a separable differential equation. We have $\frac{dy}{dx} = -y$, so

$$\int \frac{dy}{y} = \int -dx$$

and we get get the general solution $y = e^{-x + k}$. The initial condition $y(1) = 1$ implies $k = 1$ in this case.

Answer 4

If $f'(x) = -f(x)$, then

$\frac{f'(x)}{f(x)}=-1 \implies \int \frac{f'(x)}{f(x)}dx=-\int dx \implies f(x)= e^{-(x-k)}\ $ (Here k is the integration constant)

so $e^x f(x)$ is constant.

Here, $e^x f(x) = e^k$ and set $x = 1$. Then, since $f(1) = 1$, $e f(1) =e^k $ so $f(x) = e e^{-x} = e^{1-x}$.

so, $f(x)= e^{1-x} \implies\ \ $ Option (b) is correct.