Question

The integrating factor of the linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x) \] is a solution of the differential equation:

• A. \(\dfrac{dy}{dx}-P(x)y=0\)
• B. \(\dfrac{dy}{dx}+P(x)y=0\)
• C. \(\dfrac{dy}{dx}-\dfrac{y}{x}=P(x)\)
• D. \(\dfrac{dy}{dx}+\dfrac{x}{y}=P(x)\)

Hint:

The integrating factor of \[ \frac{dy}{dx}+P(x)y=Q(x) \] is always \[ e^{\int P(x)\,dx}. \] Differentiate it directly to identify the corresponding differential equation.

Answer 1

The Correct Option is A

Step 1: Recall the integrating factor formula.
For the linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] the integrating factor is \[ I.F.=e^{\int P(x)\,dx} \] Let \[ y=e^{\int P(x)\,dx} \]

Step 2: Differentiate the integrating factor.
Differentiating, \[ \frac{dy}{dx} = P(x)e^{\int P(x)\,dx} \] Since \[ y=e^{\int P(x)\,dx}, \] we get \[ \frac{dy}{dx}=P(x)y \] Rearranging, \[ \frac{dy}{dx}-P(x)y=0 \]

Step 3: Final conclusion.
Hence, the integrating factor satisfies \[ \boxed{ \frac{dy}{dx}-P(x)y=0 } \]