Q97 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
The general solution of the differential equation $(x+y)\dfrac{dy}{dx} = 1$ is
- A$x + y + 1 = c$, where $c$ is a constant of integration
- Correct answerB$x + y + 1 = ce^y$, where $c$ is a constant of integration
- C$x + y + 1 = ce^{-y}$, where $c$ is a constant of integration
- D$x + y - 1 = ce^{-y}$, where $c$ is a constant of integration
Solution
Given differential equation is $(x+y)\dfrac{dy}{dx} = 1$
Rewriting the equation, we get $\dfrac{dx}{dy} = x + y$
$\dfrac{dx}{dy} - x = y$
This is a linear differential equation of the form $\dfrac{dx}{dy} + Px = Q$, where $P = -1$ and $Q = y$.
Integrating Factor (IF) $= e^{\int P dy} = e^{\int -1 dy} = e^{-y}$
The general solution is given by $x \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dy + c$
$x e^{-y} = \int y e^{-y} dy + c$
Using integration by parts,
$x e^{-y} = y(-e^{-y}) - \int 1 \cdot (-e^{-y}) dy + c$
$x e^{-y} = -y e^{-y} - e^{-y} + c$
Multiplying both sides by $e^y$, we get
$x = -y - 1 + c e^y$
$x + y + 1 = c e^y$
Answer: $x + y + 1 = ce^y$, where $c$ is a constant of integration
Rewriting the equation, we get $\dfrac{dx}{dy} = x + y$
$\dfrac{dx}{dy} - x = y$
This is a linear differential equation of the form $\dfrac{dx}{dy} + Px = Q$, where $P = -1$ and $Q = y$.
Integrating Factor (IF) $= e^{\int P dy} = e^{\int -1 dy} = e^{-y}$
The general solution is given by $x \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dy + c$
$x e^{-y} = \int y e^{-y} dy + c$
Using integration by parts,
$x e^{-y} = y(-e^{-y}) - \int 1 \cdot (-e^{-y}) dy + c$
$x e^{-y} = -y e^{-y} - e^{-y} + c$
Multiplying both sides by $e^y$, we get
$x = -y - 1 + c e^y$
$x + y + 1 = c e^y$
Answer: $x + y + 1 = ce^y$, where $c$ is a constant of integration
Hence, the correct answer is option: 'B'.