Question

If the slope of the tangent drawn at any point \((x, y)\) on a curve is \(x + y\), then the equation of that curve is

• A. \(y = c e^x + 1 + x\)
• B. \(y = c e^{-x} - x - 1\)
• C. \(y = c e^{-x} - 1 - x\)
• D. \(y = c e^x - x - 1\)

Hint:

Solve \(\frac{dy}{dx} = x + y\) using the substitution \(v = x + y\) or as a linear differential equation with integrating factor \(e^{-x}\).

Answer 1

The Correct Option is D

The slope of the tangent is given by: \[ \frac{dy}{dx} = x + y \] Rewrite as a first-order linear differential equation: \[ \frac{dy}{dx} - y = x \] Integrating factor: \(e^{\int -1 \, dx} = e^{-x}\). Multiply through: \[ e^{-x} \frac{dy}{dx} - e^{-x} y = x e^{-x} \] \[ \frac{d}{dx} (y e^{-x}) = x e^{-x} \] Integrate: \[ y e^{-x} = \int x e^{-x} \, dx \] Use integration by parts (\(u = x\), \(dv = e^{-x} dx\)): \[ \int x e^{-x} \, dx = x (-e^{-x}) - \int (-e^{-x}) \, dx = -x e^{-x} - e^{-x} + c_1 \] \[ y e^{-x} = -x e^{-x} - e^{-x} + c_1 \] \[ y = -x - 1 + c_1 e^x = c e^x - x - 1 \] Alternatively, substitute \(v = x + y\): \[ \frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + v \] \[ \frac{dv}{1 + v} = dx \] \[ \ln |1 + v| = x + c_2 \] \[ 1 + v = c e^x \implies x + y = c e^x - 1 \implies y = c e^x - x - 1 \] Option (4) is correct. Options (1), (2), and (3) do not satisfy the differential equation.