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 $y (x) = f (x) + \int_{0}$ ˣ K(x, t)y(t) dt. Then which of the following statements are true?

A. $f^{'} (π) = 3$ ✓
B. $f (π) + f^{'} (π) = 4 - π$ ✓
C. $f (π) + f^{'} (π) = 2 - π$
D.f(0) + f′(0) = −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

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

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