Q97 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
The general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 + 5$ is ....
- Correct answerA$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$
- B$\dfrac{x^4}{4} - \dfrac{5x^2}{2} - xy = c$
- C$\dfrac{x^4}{4} - \dfrac{5x^2}{2} + xy = c$
- D$\dfrac{x^4}{4} + \dfrac{5x^2}{2} + xy = c$
Solution
The given differential equation is $\dfrac{dy}{dx} + \dfrac{1}{x}y = x^2 + 5$.
This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = \dfrac{1}{x}$ and $Q = x^2 + 5$.
The integrating factor is $e^{\int P dx} = e^{\int \dfrac{1}{x} dx} = e^{\ln x} = x$.
Multiplying the equation by the integrating factor, the solution is given by:
$y \cdot x = \int x(x^2 + 5) dx + C$
$xy = \int (x^3 + 5x) dx + C$
$xy = \dfrac{x^4}{4} + \dfrac{5x^2}{2} + C$
Rearranging the terms, we get:
$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = -C$
Taking $-C = c$, the general solution is:
$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$
Answer: $\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$
This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = \dfrac{1}{x}$ and $Q = x^2 + 5$.
The integrating factor is $e^{\int P dx} = e^{\int \dfrac{1}{x} dx} = e^{\ln x} = x$.
Multiplying the equation by the integrating factor, the solution is given by:
$y \cdot x = \int x(x^2 + 5) dx + C$
$xy = \int (x^3 + 5x) dx + C$
$xy = \dfrac{x^4}{4} + \dfrac{5x^2}{2} + C$
Rearranging the terms, we get:
$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = -C$
Taking $-C = c$, the general solution is:
$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$
Answer: $\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$
Hence, the correct answer is option: 'A'.