Question

By multiplying with $e^{\int P dx}$ on both sides of the equation $\frac{dy}{dx} + P(x)y = Q(x)$, the left side of the equation takes the form $\frac{d}{dx} (y f(x))$, then $f(x) =$

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

Hint:

The integrating factor $e^{\int P(x) \, dx}$ transforms a linear differential equation into a form where the left-hand side becomes the derivative of a product, making it easier to integrate both sides.

Answer 1

The Correct Option is C

The given differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$. The integrating factor (IF) for this equation is $e^{\int P(x) \, dx}$. Multiply both sides of the equation by the integrating factor:
\[ \begin{align} e^{\int P(x) \, dx} \frac{dy}{dx} + e^{\int P(x) \, dx} P(x)y = e^{\int P(x) \, dx} Q(x) \] The problem states that the left-hand side takes the form $\frac{d}{dx} (y f(x))$. Let’s compute the left-hand side: \[ e^{\int P(x) \, dx} \frac{dy}{dx} + e^{\int P(x) \, dx} P(x)y \] Notice that this resembles the product rule for differentiation. The product rule states that $\frac{d}{dx} (u v) = u \frac{dv}{dx} + v \frac{du}{dx}$. Let’s identify $u$ and $v$ such that the left-hand side matches this form:
- Let $v = y$ and $u = e^{\int P(x) \, dx}$.
- Then, $\frac{du}{dx} = e^{\int P(x) \, dx} P(x)$ (since the derivative of $\int P(x) \, dx$ with respect to $x$ is $P(x)$).
Applying the product rule: \[ \begin{align} \frac{d}{dx} (y e^{\int P(x) \, dx}) = e^{\int P(x) \, dx} \frac{dy}{dx} + y e^{\int P(x) \, dx} P(x) \] This matches the left-hand side exactly: \[ \begin{align} e^{\int P(x) \, dx} \frac{dy}{dx} + e^{\int P(x) \, dx} P(x)y = \frac{d}{dx} (y e^{\int P(x) \, dx}) \] Thus, the left-hand side is $\frac{d}{dx} (y f(x))$, where $f(x) = e^{\int P(x) \, dx}$. Comparing with the options, this corresponds to option (3). Thus, the correct answer is (3).