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Part BCSIR NET December 2025a-forcing-term-with-no-x-dependence-adds-an-unbounded-t-only-particular-solution

A forcing term with no x dependence adds an unbounded t only particular solution

Let u(x,t) be the solution of the initial value problem utt−uxx=e(−t),x∈R,t>0,u(x,0)=cos⁡x,ut(x,0)=0,x∈Ru_{tt} - u_{xx} = e^(-t), x \in \mathbb{R}, t > 0, u(x,0) = \cos x, u_{t}(x,0) = 0, x \in \mathbb{R}. Then which of the following statements is true?

  1. A.For each x0∈R,et⋅u(x0,t)→0x_{0} \in \mathbb{R}, e^t\cdot{}u(x_{0},t) \to 0 as t→∞t \to \infty.
  2. B.For each x0∈Rx_{0} \in \mathbb{R}, |u(x0,t)u(x_{0},t)| → 0 as t→∞t \to \infty.
  3. C.For each x0∈Rx_{0} \in \mathbb{R}, there exists at0>0a t_{0} > 0 such that u(x0,t0)≥4u(x_{0},t_{0}) \ge 4.✓ Correct
  4. D.For each x0∈Rx_{0} \in \mathbb{R}, there exists at0>0a t_{0} > 0 such that u(x0,t0)≤−4u(x_{0},t_{0}) \le -4.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: All the classical equations satisfy a maximum principle

More on this topic

The chapter behind this: The three classical equations — free to read

From Partial Differential Equations › Laplace, heat and wave equations: separation of variables

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