Assertion (A): For x > -1, the integrating factor of dy/dx - y/(x + 1) = 1 is x + 1. Reason (R): For x > -1, e^-log(x + 1) = 1/(x + 1).

Question 83

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Assertion (A): For $x > -1$ , the integrating factor of $\frac{dy}{dx} - \frac{y}{x + 1} = 1$ is $x + 1$ .
Reason (R): For $x > -1$ , $e^{-\log(x + 1)} = \frac{1}{x + 1}$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$P = -\frac{1}{x + 1}$ , so $\int P\,dx = -\log(x + 1)$ and the factor is $e^{-\log(x + 1)} = \frac{1}{x + 1}$ .
So A is false and R is true; R in fact gives the correct factor.

Worth noting

$x + 1$ is the factor for $+\frac{y}{x + 1}$ ; the minus sign inverts it.

Concept: The integrating factor
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsLinear equations and the integrating factor

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

Assertion (A): For x>−1x > -1x>−1, the integrating factor of dydx−yx+1=1\frac{dy}{dx} - \frac{y}{x + 1} = 1dxdy​−x+1y​=1 is x+1x + 1x+1.Reason (R): For x>−1x > -1x>−1, e−log⁡(x+1)=1x+1e^{-\log(x + 1)} = \frac{1}{x + 1}e−log(x+1)=x+11​.

Explanation

Assertion (A): For $x > -1$ , the integrating factor of $\frac{dy}{dx} - \frac{y}{x + 1} = 1$ is $x + 1$ .
Reason (R): For $x > -1$ , $e^{-\log(x + 1)} = \frac{1}{x + 1}$ .