Question

By multiplying with \[ e^{\int Pdx} \] on both sides of the equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] the left side of the equation turns in the form \[ \frac{d}{dx}\left(yf(x)\right), \] then \[ f(x)= \] is:

• A. \(\int y e^{\int Pdx}\,dx\)
• B. \(yP(x)\)
• C. \(e^{\int Pdx}\)
• D. \(P(x)e^{\int Pdx}\)

Hint:

For the linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] the integrating factor is always \[ e^{\int P(x)\,dx}. \]

Answer 1

The Correct Option is C

Step 1: Recall the standard linear differential equation.
The given equation is \[ \frac{dy}{dx}+P(x)y=Q(x) \] For this type of first-order linear differential equation, the integrating factor is \[ e^{\int P(x)\,dx} \]

Step 2: Multiply both sides by the integrating factor.
Multiplying by \[ e^{\int P(x)\,dx}, \] we get \[ e^{\int P(x)\,dx}\frac{dy}{dx}+P(x)y e^{\int P(x)\,dx} = Q(x)e^{\int P(x)\,dx} \]

Step 3: Express the left side as a derivative.
Using product rule, \[ \frac{d}{dx}\left(ye^{\int P(x)\,dx}\right) = e^{\int P(x)\,dx}\frac{dy}{dx} + y\frac{d}{dx}\left(e^{\int P(x)\,dx}\right) \] Now, \[ \frac{d}{dx}\left(e^{\int P(x)\,dx}\right) = P(x)e^{\int P(x)\,dx} \] Therefore, \[ \frac{d}{dx}\left(ye^{\int P(x)\,dx}\right) = e^{\int P(x)\,dx}\frac{dy}{dx} + P(x)y e^{\int P(x)\,dx} \] So the left side becomes \[ \frac{d}{dx}\left(ye^{\int P(x)\,dx}\right) \]

Step 4: Identify \(f(x)\).
Comparing \[ \frac{d}{dx}\left(yf(x)\right) \] with \[ \frac{d}{dx}\left(ye^{\int P(x)\,dx}\right), \] we get \[ f(x)=e^{\int P(x)\,dx} \]

Step 5: Final conclusion.
Hence, \[ \boxed{e^{\int Pdx}} \]