Which differential equation does the family $y = e^{x}(A + Bx)$ satisfy, $A, B$ being arbitrary constants?

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$ye^{-x} = A + Bx$ . Differentiate twice to remove $A$ , then $B$ : $\frac{d^2}{dx^2}\left(ye^{-x}\right) = 0$ .
That is $e^{-x}\left(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y\right) = 0$ , so $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$ .

Worth noting

$\frac{d^2y}{dx^2} - y = 0$ is the equation of $Ae^{x} + Be^{-x}$ ; the repeated $e^{x}$ gives $-2\frac{dy}{dx}$ .

Ten questions from Differential Equations, at your own pace.

Which differential equation does the family y=ex(A+Bx)y = e^{x}(A + Bx)y=ex(A+Bx) satisfy, A,BA, BA,B being arbitrary constants?