Question:
Integrating factor of the equation:
\[
(x^2 + 1) \frac{dy}{dx} + 2xy = x^2 - 1
\]
is:
Integrating factor of the equation:
\[
(x^2 + 1) \frac{dy}{dx} + 2xy = x^2 - 1
\]
is:
Show Hint
The integrating factor is often obtained by exponentiating the integral of the coefficient of \( y \) in the standard linear equation form.
Updated On: Apr 22, 2026
- \( \frac{2x}{x^2 + 1} \)
- \( \frac{x^2 - 1}{x^2 + 1} \)
- \( x^2 + 1 \)
- None of these
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The Correct Option is C
Solution and Explanation
Step 1: Understand the standard form for a first-order linear equation.
The given equation is in the form: \[ (x^2 + 1) \frac{dy}{dx} + 2xy = x^2 - 1 \] This is a first-order linear ordinary differential equation. To solve this, we need to find the integrating factor (I.F.), which is given by the formula: \[ \text{I.F.} = e^{\int P(x) dx} \] where \( P(x) \) is the coefficient of \( y \) in the equation after rewriting it in standard form: \[ \frac{dy}{dx} + P(x) y = Q(x) \]
Step 2: Rewrite the equation in standard form.
Divide through by \( x^2 + 1 \) to put the equation in standard form: \[ \frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{x^2 - 1}{x^2 + 1} \] Now, the coefficient of \( y \) is \( \frac{2x}{x^2 + 1} \), so: \[ P(x) = \frac{2x}{x^2 + 1} \]
Step 3: Find the integrating factor.
The integrating factor is: \[ \text{I.F.} = e^{\int \frac{2x}{x^2 + 1} dx} \] The integral of \( \frac{2x}{x^2 + 1} \) is \( \ln(x^2 + 1) \), so: \[ \text{I.F.} = e^{\ln(x^2 + 1)} = x^2 + 1 \]
Step 4: Conclusion.
Thus, the integrating factor for the given differential equation is \( x^2 + 1 \).
The given equation is in the form: \[ (x^2 + 1) \frac{dy}{dx} + 2xy = x^2 - 1 \] This is a first-order linear ordinary differential equation. To solve this, we need to find the integrating factor (I.F.), which is given by the formula: \[ \text{I.F.} = e^{\int P(x) dx} \] where \( P(x) \) is the coefficient of \( y \) in the equation after rewriting it in standard form: \[ \frac{dy}{dx} + P(x) y = Q(x) \]
Step 2: Rewrite the equation in standard form.
Divide through by \( x^2 + 1 \) to put the equation in standard form: \[ \frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{x^2 - 1}{x^2 + 1} \] Now, the coefficient of \( y \) is \( \frac{2x}{x^2 + 1} \), so: \[ P(x) = \frac{2x}{x^2 + 1} \]
Step 3: Find the integrating factor.
The integrating factor is: \[ \text{I.F.} = e^{\int \frac{2x}{x^2 + 1} dx} \] The integral of \( \frac{2x}{x^2 + 1} \) is \( \ln(x^2 + 1) \), so: \[ \text{I.F.} = e^{\ln(x^2 + 1)} = x^2 + 1 \]
Step 4: Conclusion.
Thus, the integrating factor for the given differential equation is \( x^2 + 1 \).
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