MHTCET 16 Apr 2026 Shift 2 Paper

Concept: Convert the given slope into a linear differential equation in $x$ and solve using an integrating factor.Explanation:The slope of the tangent at $P(x,y)$ is $\frac{dy}{dx} = \frac{y}{x+y}$.Since the denominator contains $x+y$, write the equation in terms of $\frac{dx}{dy}$.Taking reciprocal gives $\frac{dx}{dy} = \frac{x+y}{y} = \frac{x}{y} + 1$.Rearrange to get $\frac{dx}{dy} - \frac{x}{y} = 1$.This is a linear differential equation in $x$.Its integrating factor is $e^{\int -\frac{1}{y}dy} = e^{-\ln y} = \frac{1}{y}$.Multiplying throughout by $\frac{1}{y}$, we obtain $\frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} = \frac{1}{y}$.The left side is $\frac{d}{dy}\left(\frac{x}{y}\right)$, so $\frac{d}{dy}\left(\frac{x}{y}\right) = \frac{1}{y}$.Integrating both sides: $\frac{x}{y} = \ln y + C$.The curve passes through $(0,1)$, so put $x=0$, $y=1$: $0 = \ln 1 + C$ gives $C=0$.Thus $\frac{x}{y} = \ln y$, which in exponential form is $y = e^{x/y}$.Answer: Option D: $y = e^{\frac{x}{y}}$