Question

In the following differential equation, the numerically obtained value of $y(t)$ at $t=1$ is \underline{\hspace{3cm}} (Round off to 2 decimal places). \[ \frac{dy}{dt} = \frac{e^{-\alpha t}}{2 + \alpha t}, \alpha = 0.01, \, y(0)=0 \]

Hint:

For small $\alpha$, use series expansion or approximation to simplify integrals. Here, denominator variation was negligible compared to numerator decay.

Answer 1

Step 1: Integrate expression.
\[ y(t) = \int_{0}^{t} \frac{e^{-\alpha \tau}}{2 + \alpha \tau} \, d\tau \] At $t=1$, $\alpha=0.01$: \[ y(1) = \int_{0}^{1} \frac{e^{-0.01\tau}}{2 + 0.01\tau} \, d\tau \]

Step 2: Approximate denominator.
Since $0.01\tau \ll 2$, denominator $\approx 2.00$ to $2.01$. So: \[ y(1) \approx \frac{1}{2} \int_{0}^{1} e^{-0.01\tau} \, d\tau \]

Step 3: Solve integral.
\[ \int_{0}^{1} e^{-0.01\tau} d\tau = \left[\frac{-1}{0.01} e^{-0.01\tau}\right]_{0}^{1} = -100(e^{-0.01}-1) \] \[ = 100(1 - 0.99005) = 0.995 \] \[ y(1) \approx \frac{0.995}{2} = 0.497 \] Rounding off: $0.49$. % Final Answer \[ \boxed{0.49} \]