What is the integrating factor of (1 - x^2)dy/dx - xy = 1, for -1 < x < 1?

Question 89

HardOriginal question

What is the integrating factor of $(1 - x^{2})\frac{dy}{dx} - xy = 1$ , for $-1 < x < 1$ ?

\sqrt{1 - x^{2}}

Correct answerYour answer

\frac{1}{\sqrt{1 - x^{2}}}

Correct answerYour answer

1 - x^{2}

Correct answerYour answer

e^{- \frac{x^{2}}{2}}

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Divide by $1 - x^{2}$ : $P = -\frac{x}{1 - x^{2}}$ .
$\int -\frac{x}{1 - x^{2}}\,dx = \frac{1}{2}\log(1 - x^{2})$ , so the factor is $\sqrt{1 - x^{2}}$ .

Worth noting

$\frac{1}{\sqrt{1 - x^{2}}}$ gets the sign of the logarithm wrong: the derivative of $1 - x^{2}$ is $-2x$ , which absorbs the minus in $P$ .

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