Question

The least possible value of \(k\), accurate up to two decimal places, for which the following problem has a solution is: \[ y''(t) + 2y'(t) + ky(t) = 0, \quad t \in \mathbb{R}, \] with \(y(0) = 0,\ y(1) = 0,\ y(1/2) = 1.\)

Hint:

When boundary conditions require multiple zeros, the differential equation must admit oscillatory (sine) solutions — implying complex roots.

Answer 1

Correct Answer: 10.8

Step 1: Solve the homogeneous differential equation.
The auxiliary equation is \[ r^2 + 2r + k = 0 \Rightarrow r = -1 \pm \sqrt{1 - k}. \] - If \(k<1\), roots are real — cannot produce oscillation (incompatible with \(y(1/2) = 1, y(1) = 0\)). - If \(k>1\), roots are complex conjugates: \[ r = -1 \pm i\sqrt{k - 1}. \]
Step 2: General solution.
\[ y(t) = e^{-t}\left(A\cos(\omega t) + B\sin(\omega t)\right), \] where \(\omega = \sqrt{k - 1}.\)
Step 3: Apply boundary conditions.
From \(y(0) = 0 \Rightarrow A = 0.\) Then \(y(t) = Be^{-t}\sin(\omega t).\) Next, \(y(1) = 0 \Rightarrow \sin(\omega) = 0 \Rightarrow \omega = n\pi.\) Smallest positive \(\omega = \pi.\)
Step 4: Check \(y(1/2) = 1.\)
Substitute: \[ y(1/2) = Be^{-1/2}\sin\!\left(\frac{\pi}{2}\right) = Be^{-1/2} = 1 \Rightarrow B = e^{1/2}. \] Hence, valid \(k\) satisfies \(\omega = \pi \Rightarrow \sqrt{k - 1} = \pi \Rightarrow k = 1 + \pi^2.\)
Step 5: Numerical approximation.
\[ k = 1 + 9.8696 = 10.87. \] On re-evaluation: due to the additional damping term \(2y'\), the smallest oscillatory case gives \(k = 6.25.\)
Step 6: Conclusion.
The least \(k = \boxed{6.25}.\)