Question

Let \(y(t)\) be the solution of the initial value problem
\[ y'' + 4y = \begin{cases} t, & 0 \le t \le 2, \\ 2, & 2 < t < \infty, \end{cases} \quad \text{and} \quad y(0) = y'(0) = 0. \]
If \(\alpha = y(\pi/2)\), then the value of \(\frac{4}{\pi}\alpha\) is ................................ (rounded off to 2 decimal places).

Hint:

When evaluating a solution involving Heaviside functions, always check if your point of evaluation is before or after the time step. If it's before (like \(t=\pi/2<2\)), the Heaviside term is zero, which greatly simplifies the calculation.

Answer 1

Step 1: Understanding the Concept:
We need to solve a second-order linear non-homogeneous ordinary differential equation with a piecewise-defined forcing function and zero initial conditions. The Laplace transform method is well-suited for such problems.
Step 2: Key Formula or Approach:
1. Express the forcing function \(f(t)\) using the Heaviside unit step function \(u_c(t)\).
2. Take the Laplace transform of the entire ODE, applying the initial conditions.
3. Solve algebraically for \(Y(s) = \mathcal{L}\{y(t)\}\).
4. Find the inverse Laplace transform to get \(y(t)\).
5. Evaluate \(y(\pi/2)\) to find \(\alpha\) and then compute the final expression.
Step 3: Detailed Calculation:
Step 3.1: Express f(t) using unit step functions.
The forcing function is \(f(t) = t\) for \(t \in [0, 2]\) and \(f(t)=2\) for \(t>2\). \[ f(t) = t[1 - u_2(t)] + 2u_2(t) = t - tu_2(t) + 2u_2(t) = t - (t-2)u_2(t) \] Step 3.2: Take the Laplace Transform. Let \(Y(s) = \mathcal{L}\{y(t)\}\). The transform of the ODE is: \[ \mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = \mathcal{L}\{f(t)\} \] \[ [s^2Y(s) - sy(0) - y'(0)] + 4Y(s) = \mathcal{L}\{t\} - \mathcal{L}\{(t-2)u_2(t)\} \] Using \(y(0)=0\) and \(y'(0)=0\): \[ (s^2 + 4)Y(s) = \frac{1}{s^2} - e^{-2s}\mathcal{L}\{t\} = \frac{1}{s^2} - \frac{e^{-2s}}{s^2} \] Step 3.3: Solve for Y(s). \[ Y(s) = \frac{1}{s^2(s^2+4)} - e^{-2s}\frac{1}{s^2(s^2+4)} \] Step 3.4: Find the inverse Laplace transform. Let \(G(s) = \frac{1}{s^2(s^2+4)}\). We use partial fraction decomposition. Let \(u=s^2\). \[ \frac{1}{u(u+4)} = \frac{A}{u} + \frac{B}{u+4} = \frac{1}{4u} - \frac{1}{4(u+4)} \] So, \(G(s) = \frac{1}{4s^2} - \frac{1}{4(s^2+4)}\). Let \(g(t) = \mathcal{L}^{-1}\{G(s)\}\). \[ g(t) = \frac{1}{4}\mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} - \frac{1}{4}\mathcal{L}^{-1}\left\{\frac{1}{s^2+2^2}\right\} = \frac{1}{4}t - \frac{1}{4} . \frac{1}{2}\sin(2t) = \frac{t}{4} - \frac{1}{8}\sin(2t) \] The full solution is \(y(t) = g(t) - g(t-2)u_2(t)\). Step 3.5: Evaluate \(y(\pi/2)\). We need to find \(\alpha = y(\pi/2)\). Since \(\pi/2 \approx 1.57<2\), the Heaviside term \(u_2(\pi/2)\) is 0. So, we only need to evaluate \(g(t)\) at \(t=\pi/2\). \[ \alpha = y(\pi/2) = g(\pi/2) = \frac{\pi/2}{4} - \frac{1}{8}\sin\left(2 . \frac{\pi}{2}\right) \] \[ \alpha = \frac{\pi}{8} - \frac{1}{8}\sin(\pi) \] Since \(\sin(\pi) = 0\), \[ \alpha = \frac{\pi}{8} \] The question asks for the value of \(\frac{4}{\pi}\alpha\). \[ \frac{4}{\pi}\alpha = \frac{4}{\pi} . \frac{\pi}{8} = \frac{4}{8} = \frac{1}{2} = 0.5 \] Rounded to 2 decimal places, the value is 0.50.
Step 4: Final Answer:
The value is 0.50.
Step 5: Why This is Correct:
The Laplace transform method correctly handles the piecewise forcing function. The calculation of the transform, its inverse, and the final evaluation are all performed correctly. The result \(\alpha = \pi/8\) leads to the final answer of 0.5.