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Part CCSIR NET June 2025integrating-the-ode-twice-puts-the-initial-data-into-f-not-into-the-kernel

Integrating the ode twice puts the initial data into f not into the kernel

Let f and K be such that the solution of the initial value problem y″ − 3y′ + 2y = 4sin(x), y(0) = 1, y′(0) = −2 satisfies the Volterra integral equation y(x)=f(x)+∫0y(x) = f(x) + \int_{0}ˣ K(x, t)y(t) dt. Then which of the following statements are true?

  1. A.f′(π)=3f'(\pi) = 3✓ Correct
  2. B.f(π)+f′(π)=4−πf(\pi) + f'(\pi) = 4 - \pi✓ Correct
  3. C.f(π)+f′(π)=2−πf(\pi) + f'(\pi) = 2 - \pi
  4. D.f(0) + f′(0) = −4✓ Correct

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: Every integral equation of the second kind has a unique solution

More on this topic

The chapter behind this: Integral equations: Fredholm vs Volterra — free to read

From Linear Integral Equations › Fredholm and Volterra equations

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