Question

The integrating factor of following differential equation is : $\frac{dy}{dx}-\frac{3x^{2}-1}{x^{3}-x}y=x^{2}-1$}

• A. $\frac{1}{x^{2}-1}$
• B. $\frac{1}{log(x^{3}-x)}$
• C. $\frac{1}{x^{3}-x}$
• D. $\frac{3x^{2}-1}{x^{3}-x}$

Hint:

Whenever the integrand is of the form $f'(x)/f(x)$, the integral is simply $\ln(f(x))$. This pattern appears very frequently in integrating factor problems.

Answer 1

The Correct Option is C

Concept: A first-order linear differential equation has the form $\frac{dy}{dx} + P(x)y = Q(x)$. The Integrating Factor (I.F.) is calculated using the formula: $I.F. = e^{\int P(x) dx}$

Step 1: Identify $P(x)$.
In the given equation $\frac{dy}{dx} - \frac{3x^2-1}{x^3-x}y = x^2-1$: $P(x) = -\frac{3x^2-1}{x^3-x}$

Step 2: Integrate $P(x)$.
$\int P(x) dx = \int -\frac{3x^2-1}{x^3-x} dx$. Notice that the numerator $(3x^2-1)$ is exactly the derivative of the denominator $(x^3-x)$. Using $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|$: $\int P(x) dx = -\ln(x^3-x) = \ln(x^3-x)^{-1} = \ln(\frac{1}{x^3-x})$.

Step 3: Calculate the Exponential.
$I.F. = e^{\ln(\frac{1}{x^3-x})} = \frac{1}{x^3-x}$. This matches option (3).