MHTCET 13 Apr 2026 Shift 1 Paper

Concept:This is a homogeneous differential equation, solved by substituting $y = vx$.Explanation:Given: $\frac{dy}{dx} = \frac{x+y}{x-y}$.Both numerator and denominator have degree 1, so the equation is homogeneous.Put $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substitute in the equation:$v + x\frac{dv}{dx} = \frac{1+v}{1-v}$.Therefore, $x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}$.Separating variables:$\frac{1-v}{1+v^2}\,dv = \frac{dx}{x}$.Integrate both sides:$\int \frac{1-v}{1+v^2}\,dv = \int \frac{dx}{x}$.This gives $\tan^{-1}v - \frac{1}{2}\ln(1+v^2) = \ln|x| + C$.Substitute $v = \frac{y}{x}$:$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln\left(1+\frac{y^2}{x^2}\right) = \ln|x| + C$.Since $1+\frac{y^2}{x^2} = \frac{x^2+y^2}{x^2}$, we get:$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln(x^2+y^2) = C$.Writing in exponential form:$\frac{e^{\tan^{-1}(y/x)}}{(x^2+y^2)^{1/2}} = C$, or $C\,(x^2+y^2)^{1/2} = e^{\tan^{-1}(y/x)}$.Replacing $C$ by an arbitrary constant $c$,$c\left(x^2+y^2\right)^{1/2} = e^{\tan^{-1}\left(\frac{y}{x}\right)}$.Answer:Option B.