Discussion
The solution of the integral equation :
\[ x(t) = e^{-t} + \int_{0}^{t} \sin(t - \tau) x(\tau) d\tau \]
by using Laplace transform method is :
Answer: D
The solution is \(x(t) = 2e^{-t} + t - 1\).
The integral is the convolution \(\sin t * x(t)\). Taking Laplace transforms, with \(L\{\sin t\} = 1/(s^{2} + 1)\):
\[ \begin{aligned} X &= \frac{1}{s + 1} + \frac{X}{s^{2} + 1} \\ X\frac{s^{2}}{s^{2} + 1} &= \frac{1}{s + 1} \\ X &= \frac{s^{2} + 1}{s^{2}(s + 1)} \end{aligned} \]
Partial fractions give \(X = -\tfrac{1}{s} + \tfrac{1}{s^{2}} + \tfrac{2}{s + 1}\), so \(x(t) = -1 + t + 2e^{-t}\). Check: \(x(0) = 1\), which matches \(e^{0} + 0\). The options with \(e^{-2t}\) have the wrong exponential, and \(e^{-t} + 2t - 1\) has \(x'(0) = 1\), while differentiating the equation gives \(x'(0) = -1\).
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