Q65 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { Integrating factor of the differential equation } \dfrac{d y}{d x}+y=\dfrac{x^3+y}{x} \text { is }$
- A$x e^x$
- Correct answerB$\dfrac{e^x}{x}$
- C$\dfrac{x}{e^x}$
- D$e^x$
Solution
The given differential equation is $\dfrac{dy}{dx} + y = \dfrac{x^3 + y}{x}$.
Rearranging the terms to the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$, we have:
$\dfrac{dy}{dx} + y = x^2 + \dfrac{y}{x}$
$\dfrac{dy}{dx} + y - \dfrac{y}{x} = x^2$
$\dfrac{dy}{dx} + y\left(1 - \dfrac{1}{x}\right) = x^2$
Here, $P(x) = 1 - \dfrac{1}{x}$.
The integrating factor (IF) is given by $e^{\int P(x) dx}$.
IF $= e^{\int (1 - \dfrac{1}{x}) dx} = e^{x - \ln|x|} = e^x \cdot e^{-\ln|x|} = e^x \cdot \dfrac{1}{x} = \dfrac{e^x}{x}$.
Answer: $\dfrac{e^x}{x}$
Rearranging the terms to the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$, we have:
$\dfrac{dy}{dx} + y = x^2 + \dfrac{y}{x}$
$\dfrac{dy}{dx} + y - \dfrac{y}{x} = x^2$
$\dfrac{dy}{dx} + y\left(1 - \dfrac{1}{x}\right) = x^2$
Here, $P(x) = 1 - \dfrac{1}{x}$.
The integrating factor (IF) is given by $e^{\int P(x) dx}$.
IF $= e^{\int (1 - \dfrac{1}{x}) dx} = e^{x - \ln|x|} = e^x \cdot e^{-\ln|x|} = e^x \cdot \dfrac{1}{x} = \dfrac{e^x}{x}$.
Answer: $\dfrac{e^x}{x}$
Hence, the correct answer is option: 'B'.