For x > 0, what is the general solution of xdy/dx - y = x^2e^x?

Homogeneous & Linear Differential Equations · Question 109 of 120

Question 109

MediumOriginal question

For $x > 0$ , what is the general solution of $x\frac{dy}{dx} - y = x^{2}e^{x}$ ?

y = e^{x} + Cx

Correct answerYour answer

y = xe^{x} + C

Correct answerYour answer

y = x^{2}e^{x} + Cx

Correct answerYour answer

y = xe^{x} + Cx

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Divide by $x$ : $\frac{dy}{dx} - \frac{1}{x}y = xe^{x}$ , with factor $\frac{1}{x}$ .
$\frac{y}{x} = \int e^{x}\,dx = e^{x} + C$ , so $y = xe^{x} + Cx$ .

Worth noting

$e^{x} + Cx$ multiplies only the constant by $x$ at the end; $e^{x}$ must be multiplied too.

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

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