Question

The integrating factor of the differential equation \( x^2 \left(\frac{dy}{dx}\right) - xy = 1 \) is:

• A. \( 1/x \)
• B. \( \frac{x}{1+x^2} \)
• C. \( \sqrt{1-x^2} \)
• D. None of these

Hint:

To find the integrating factor for a first-order linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \), first ensure the equation is in this standard form. Then, calculate the integrating factor using \( \text{IF} = e^{\int P(x) dx} \). Remember properties of logarithms and exponentials, like \( e^{\ln u} = u \) and \( a \ln b = \ln(b^a) \).

Answer 1

The Correct Option is A

Step 1: Recognize the given differential equation and convert it into the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \). The given equation is: \[ x^2 \frac{dy}{dx} - xy = 1 \] Divide the entire equation by \( x^2 \) (assuming \( x \neq 0 \)) to make the coefficient of \( \frac{dy}{dx} \) equal to 1: \[ \frac{dy}{dx} - \frac{xy}{x^2} = \frac{1}{x^2} \] \[ \frac{dy}{dx} - \left(\frac{1}{x}\right)y = \frac{1}{x^2} \] Step 2: Identify the functions \( P(x) \) and \( Q(x) \) by comparing the equation with the standard form. \[ P(x) = -\frac{1}{x} \] \[ Q(x) = \frac{1}{x^2} \] Step 3: Calculate the integrating factor (IF) using the formula \( \text{IF} = e^{\int P(x) dx} \). \[ \text{IF} = e^{\int \left(-\frac{1}{x}\right) dx} \] \[ \text{IF} = e^{-\int \frac{1}{x} dx} \] The integral of \( \frac{1}{x} \) is \( \ln|x| \). \[ \text{IF} = e^{-\ln|x|} \] Using the logarithm property \( - \ln a = \ln(a^{-1}) = \ln(1/a) \): \[ \text{IF} = e^{\ln\left|\frac{1}{x}\right|} \] Since \( e^{\ln u} = u \): \[ \text{IF} = \left|\frac{1}{x}\right| \] We typically take the positive value for the integrating factor, so: \[ \text{IF} = \frac{1}{x} \] Step 4: Compare the result with the given options. The calculated integrating factor is \( 1/x \), which matches option (A).