y = e^x + e^4x satisfies d^2y/dx^2 + pdy/dx + qy = 0. What are p and q?

Question 95

HardOriginal question

$y = e^{x} + e^{4x}$ satisfies $\frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = 0$ . What are $p$ and $q$ ?

p = 5,\ q = 4

Correct answerYour answer

p = -5,\ q = 4

Correct answerYour answer

p = -5,\ q = -4

Correct answerYour answer

p = -4,\ q = 5

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$\frac{dy}{dx} = e^{x} + 4e^{4x}$ and $\frac{d^2y}{dx^2} = e^{x} + 16e^{4x}$ .
Each exponential must vanish on its own: $1 + p + q = 0$ and $16 + 4p + q = 0$ . Subtracting, $15 + 3p = 0$ , so $p = -5$ and $q = 4$ .

Worth noting

$p = 5$ flips the sign: the powers 1 and 4 are the roots of $m^{2} + pm + q = 0$ , so $p = -(1 + 4)$ and $q = 1 \times 4$ .

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