MHT CET202616 April 2026Evening ShiftMathematicsDifferential EquationsActual
The equation of the curve passing through the point (0, 1) and having a slope of the tangent at point P(x, y) is equal to y x + y , is
Options
- Ax = e^ - y x
- By = e^ - y x + 1
- Cx = e^ x y - 1
- Dy = e^ x y
Correct answer
D. y = e^ x y
Step-by-step solution
Given the slope of the tangent is dy dx = y x + y Taking the reciprocal, we get: dx dy = x + y y dx dy - x y = 1 This is a linear differential equation of the form dx dy + Px = Q , where P = - 1 y and Q = 1 . Integrating Factor (IF) = e^ - 1 y dy = e^ - y = 1 y The solution is given by: x ( IF ) = Q ( IF ) dy + C x ( 1 y ) = 1 y dy + C x y = y + C Since the curve passes through the point (0, 1) , substitute x = 0 and y = 1 : 0 1 = 1 + C C = 0 Therefore, the equation of the curve is: x y = y y = e^ x y Answer: y = e