Q243 / 270 · NDA 2025 (Phase 2) · Mathematics
Consider the following statements:
I. $y = xe^{2x}$ is the solution of $\dfrac{dy}{dx} = y\left(2 + \dfrac{1}{x}\right)$.
II. $y = x\ln|x| + cx$ is the solution of $\dfrac{dy}{dx} = \dfrac{x + y}{x}$.
Which of the statements given above is/are correct?
I. $y = xe^{2x}$ is the solution of $\dfrac{dy}{dx} = y\left(2 + \dfrac{1}{x}\right)$.
II. $y = x\ln|x| + cx$ is the solution of $\dfrac{dy}{dx} = \dfrac{x + y}{x}$.
Which of the statements given above is/are correct?
- AI only
- BII only
- Correct answerCBoth I and II
- DNeither I nor II
Solution
For Statement I:
Given $y = xe^{2x}$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x)$
Also, $y\left(2 + \dfrac{1}{x}\right) = xe^{2x}\left(\dfrac{2x + 1}{x}\right) = e^{2x}(2x + 1)$
Since $\dfrac{dy}{dx} = y\left(2 + \dfrac{1}{x}\right)$, Statement I is correct.
For Statement II:
Given $y = x\ln|x| + cx$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \ln|x| + x\left(\dfrac{1}{x}\right) + c = \ln|x| + 1 + c$
Also, $\dfrac{x + y}{x} = 1 + \dfrac{y}{x} = 1 + \dfrac{x\ln|x| + cx}{x} = 1 + \ln|x| + c$
Since $\dfrac{dy}{dx} = \dfrac{x + y}{x}$, Statement II is correct.
Both statements I and II are correct.
Answer: Both I and II
Given $y = xe^{2x}$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x)$
Also, $y\left(2 + \dfrac{1}{x}\right) = xe^{2x}\left(\dfrac{2x + 1}{x}\right) = e^{2x}(2x + 1)$
Since $\dfrac{dy}{dx} = y\left(2 + \dfrac{1}{x}\right)$, Statement I is correct.
For Statement II:
Given $y = x\ln|x| + cx$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = \ln|x| + x\left(\dfrac{1}{x}\right) + c = \ln|x| + 1 + c$
Also, $\dfrac{x + y}{x} = 1 + \dfrac{y}{x} = 1 + \dfrac{x\ln|x| + cx}{x} = 1 + \ln|x| + c$
Since $\dfrac{dy}{dx} = \dfrac{x + y}{x}$, Statement II is correct.
Both statements I and II are correct.
Answer: Both I and II
Hence, the correct answer is option: 'C'.