$y = e^{x}(A\cos x + B\sin x)$ , with $A$ and $B$ constants. Which relation does $y$ satisfy?

\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} - 2\frac{dy}{dx} - 2y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} + y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$\frac{dy}{dx} = y + e^{x}(-A\sin x + B\cos x)$ , so $e^{x}(-A\sin x + B\cos x) = \frac{dy}{dx} - y$ .
Differentiating again: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + \left(\frac{dy}{dx} - y\right) - y$ , that is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$ .

Worth noting

$\frac{d^2y}{dx^2} + y = 0$ is the relation for $A\cos x + B\sin x$ alone; the factor $e^{x}$ adds the middle term.

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

y=ex(Acos⁡x+Bsin⁡x)y = e^{x}(A\cos x + B\sin x)y=ex(Acosx+Bsinx), with AAA and BBB constants. Which relation does yyy satisfy?