Question 13 of 96beginner🔧 ApplyShort Answer2 marks

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y=ex+1y = e^x + 1 : y′′−y′=0y'' - y' = 0

Correct Answer

The given function y = exe^{x} + 1 is a solution of the differential equation y'' - y' = 0. Given: y = exe^{x} + 1

Step 1: Find the first derivative. y' = dy/dx = d/dx(exe^{x} + 1) = exe^{x}

Step 2: Find the second derivative. y'' = d²y/dx² = d/dx(exe^{x}) = exe^{x}

Step 3: Substitute in the differential equation. LHS = y'' - y' = exe^{x} - exe^{x} = 0 = RHS

Since LHS = RHS, the given function y = exe^{x} + 1 is verified to be a solution of the differential equation y'' - y' = 0.

Exercise: EXERCISE 9.2 | Q: 1 | (Page 7)
For More Understanding

Explanation

This is Exercise 9.2, Question 1 from NCERT Class 12 Mathematics. The verification process involves differentiating the given function twice and substituting into the differential equation. The constant term '1' in the function disappears upon differentiation, which is why it satisfies the homogeneous equation y'' - y' = 0.

Solution Steps

  1. Step 1: Differentiate y = exe^{x} + 1 to get y' = exe^{x}

  2. Step 2: Differentiate y' = exe^{x} to get y'' = exe^{x}

  3. Step 3: Substitute in y'' - y' = 0 to get exe^{x} - exe^{x} = 0, which is verified

PreviousNext