Question:medium

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

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"Ordinate" is $y$ and "Abscissa" is $x$.
Updated On: Jun 19, 2026
  • $y=x-4-2e^{x}$
  • $y=4-x-2e^{x}$
  • $y=4+x-2e^{x}$
  • $y=4-x+2e^{x}$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
Translate the given physical condition into a differential equation and solve it with the given initial condition.

Step 2: Key Formula or Approach:

1. Abscissa $= x$, Ordinate $= y$, Slope $= \frac{dy}{dx}$.
2. Condition: $x + y = \frac{dy}{dx} + 5$.
3. Solve using the integrating factor (IF) method for linear differential equations.

Step 3: Detailed Explanation:

The differential equation is: \[ \frac{dy}{dx} - y = x - 5 \] This is a linear DE of the form $\frac{dy}{dx} + Py = Q$, where $P = -1$ and $Q = x - 5$.
Integrating factor $IF = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
Solution: $y \cdot IF = \int Q \cdot IF dx + C$
\[ y e^{-x} = \int (x - 5) e^{-x} dx + C \] Using integration by parts ($\int u v = u \int v - \int (u' \int v)$): \[ y e^{-x} = (x - 5)(-e^{-x}) - \int (1)(-e^{-x}) dx + C \] \[ y e^{-x} = -(x - 5)e^{-x} - e^{-x} + C = (4 - x)e^{-x} + C \] \[ y = 4 - x + C e^x \] The curve passes through $(0, 2)$: \[ 2 = 4 - 0 + C e^0 \Rightarrow 2 = 4 + C \Rightarrow C = -2 \] The equation is $y = 4 - x - 2e^x$.

Step 4: Final Answer:

The equation is $y = 4 - x - 2e^x$.
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