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

Homogeneous & Linear Differential Equations · Question 113 of 120

Question 113

HardOriginal question

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

0

Correct answerYour answer

\frac{2}{e}

Correct answerYour answer

2 - \frac{2}{e}

Correct answerYour answer

\frac{4}{e}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

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

Worth noting

0 drops the $Ce^{-x}$ term, taking $y = 2x - 2$ alone.

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

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