Question

The equation of the curve through (0,2) given the sum of ordinate and abscissa at any point exceeds the slope of tangent by 5 is

• A. $y=x-4-2e^{x}$
• B. $y=4-x-2e^{x}$
• C. $y=4+x-2e^{x}$
• D. $y=4-x+2e^{x}$

Hint:

"Ordinate" is $y$ and "Abscissa" is $x$.

Answer 1

The Correct Option is B

Step 1: Differential Equation
Sum $(y+x)$ exceeds slope $(dy/dx)$ by 5: $y + x = \frac{dy}{dx} + 5$.
$\frac{dy}{dx} - y = x - 5$.

Step 2: Integrating Factor
This is a linear DE of form $dy/dx + Py = Q$.
$P = -1, Q = x - 5$.
$I.F. = e^{\int -1 dx} = e^{-x}$.

Step 3: Calculation
$y \cdot e^{-x} = \int (x-5)e^{-x} dx = -(x-5)e^{-x} + \int e^{-x} dx = -(x-5)e^{-x} - e^{-x} + C$.
$y \cdot e^{-x} = (5-x-1)e^{-x} + C = (4-x)e^{-x} + C$.
$y = 4 - x + Ce^x$.
At $(0,2)$: $2 = 4 - 0 + C(1) \implies C = -2$.

Step 4: Conclusion
Hence, $y = 4 - x - 2e^x$. Final Answer: (B)