If \(y = e^{2x}(A\cos x + B\sin x)\), where A and B are arbitrary constants, then y satisfies the differential equation
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0\)
Let \(z = e^{2x}(-A\sin x + B\cos x)\). By the product rule:
\( \displaystyle \frac{dy}{dx} = 2e^{2x}(A\cos x + B\sin x) + e^{2x}(-A\sin x + B\cos x) \)
\( \displaystyle \frac{dy}{dx} = 2y + z \)
In the same way,
\( \displaystyle \frac{dz}{dx} = 2z + e^{2x}(-A\cos x - B\sin x) = 2z - y \)
Differentiating \(\frac{dy}{dx} = 2y + z\) again:
\( \displaystyle \frac{d^2y}{dx^2} = 2\frac{dy}{dx} + \frac{dz}{dx} = 2\frac{dy}{dx} + 2z - y \)
Eliminating z with \(z = \frac{dy}{dx} - 2y\):
\( \displaystyle \frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2\frac{dy}{dx} - 4y - y \)
\( \displaystyle \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0 \)
Why the other options are incorrect
Option (2) \(\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 0\) — Incorrect. This is satisfied by \(e^{-2x}(A\cos x + B\sin x)\); the sign of the middle term goes with the sign of the power of e.
Option (3) \(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 3y = 0\) — Incorrect. Since \(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} = -5y\), the left side equals \(-2y\), not zero.
Option (4) \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 5y = 0\) — Incorrect. Since \(\frac{d^2y}{dx^2} = 4\frac{dy}{dx} - 5y\), the left side equals \(2\frac{dy}{dx}\), not zero.