y satisfies dy/dx + y = 1 and y = 0 when x = 0. What is y when x = log 2?

Homogeneous & Linear Differential Equations · Question 98 of 120

Question 98

MediumOriginal question

$y$ satisfies $\frac{dy}{dx} + y = 1$ and $y = 0$ when $x = 0$ . What is $y$ when $x = \log 2$ ?

\frac{1}{2}

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1

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\frac{3}{2}

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2

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

The factor $e^{x}$ gives $ye^{x} = e^{x} + C$ , so $y = 1 + Ce^{-x}$ ; $y(0) = 0$ gives $C = -1$ .
$y = 1 - e^{-\log 2} = 1 - \frac{1}{2} = \frac{1}{2}$ .

Worth noting

$\frac{3}{2}$ takes $C = +1$ ; the condition $y(0) = 0$ needs $1 + C = 0$ .

Ten questions from Homogeneous & Linear Differential Equations, at your own pace.

Ten questions from Homogeneous & Linear Differential Equations, at your own pace.