The solution of $f ( t ) = \cos t + \int _ { 0 } ^ { t } e ^ { - \tau } f ( t - \tau ) d \tau$ is
A) $f ( t ) = - \cos t + \sin t$
B) $f ( t ) = \cos t - \sin t$
C) $f ( t ) = \cos t + \sin t$
D) $f ( t ) = - \cos t - \sin t$
E) $f ( t ) = \cos t - 2 \sin t$

Question

Quizplus · Accepted Answer

This equation is a form of an integral equation where $f(t)$ is defined in terms of itself, which suggests a solution involving differential equations. Differentiating both sides of the equation with respect to $t$ gives:$$f'(t) = -\sin t + e^0 f(0) = -\sin t + f(0)$$Since $f(0) = \cos(0) = 1$, we have:$$f'(t) = -\sin t + 1$$Integrating this expression to find $f(t)$, we get:$$f(t) = \cos t + \sin t + C$$Given the original equation, we can determine that $C = 0$ because the integral term accounts for the modification to the cosine term, leading to the solution:$$f(t) = \cos t + \sin t$$This matches choice C.

The Solution Of f(t)=cos⁡t+∫0te−τf(t−τ)dτf ( t ) = \cos t + \int _ { 0 } ^ { t } e ^ { - \tau } f ( t - \tau ) d \tauf(t)=cost+∫0t​e−τf(t−τ)dτ

The Solution Of $f ( t ) = \cos t + \int _ { 0 } ^ { t } e ^ { - \tau } f ( t - \tau ) d \tau$