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Part CCSIR NET December 2024the-sign-of-the-larger-root-decides-boundedness

The sign of the larger root decides boundedness

Consider the ordinary differential equation (ODE)d2y/) d^{2}y/dx2+2(e(2x)−1)(^{2} + 2(e^(2x) - 1)(dy/dx)+αe(4x)y=0,x∈R) + \alpha{}e^(4x)y = 0, x \in \mathbb{R}, where α∈R\alpha \in \mathbb{R}. Note that the ODE transforms into an equation with constant coefficients under the change of independent variable given by t = e^(2x)/2. Then which of the following statements are true?

  1. A.For α=1\alpha = 1, all the solutions of the ODE tend to zero as x→∞x \to \infty✓ Correct
  2. B.For α=0\alpha = 0, there exists a solution of the ODE which tends to 1 as x→∞x \to \infty✓ Correct
  3. C.For α=−1\alpha = -1, there exists an unbounded solution of the ODE on R\mathbb{R}✓ Correct
  4. D.For α=2\alpha = 2, there exists an unbounded solution of the ODE on R\mathbb{R}

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 boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: W(f, g) ≡ 0 implies f and g are linearly dependent

More on this topic

The chapter behind this: Linear ODE, Wronskian and systems — free to read

From Ordinary Differential Equations › Linear ODE, Wronskian, variation of parameters, systems

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