Question 14 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=x2+2x+Cy = x^2 + 2x + C : y′−2x−2=0y' - 2x - 2 = 0

Correct Answer

Given function: y=x2+2x+Cy = x^2 + 2x + C

Differentiating both sides with respect to xx:

y′=dydx=2x+2y' = \frac{dy}{dx} = 2x + 2

Substituting y′y' in the given differential equation y′−2x−2=0y' - 2x - 2 = 0: L.H.S. = y′−2x−2y' - 2x - 2 = (2x+2)−2x−2(2x + 2) - 2x - 2 = 00 = R.H.S.

Since L.H.S. = R.H.S., the given function y=x2+2x+Cy = x^2 + 2x + C is a solution of the corresponding differential equation y′−2x−2=0y' - 2x - 2 = 0.

Exercise: EXERCISE 9.2 | Q: 2 | (Page 7)
For More Understanding

Explanation

This is a verification problem from Exercise 9.2. The student needs to differentiate the given function and substitute into the differential equation. The derivative of x2x^2 is 2x2x, derivative of 2x2x is 22, and derivative of constant CC is 00. When substituted into the DE, all terms cancel to give 0, verifying the solution.

Solution Steps

  1. Step 1: Differentiate the given function y=x2+2x+Cy = x^2 + 2x + C with respect to xx to get y′=2x+2y' = 2x + 2

  2. Step 2: Substitute y′=2x+2y' = 2x + 2 into the differential equation y′−2x−2=0y' - 2x - 2 = 0

  3. Step 3: Simplify L.H.S. = (2x+2)−2x−2=0(2x + 2) - 2x - 2 = 0, which equals R.H.S., thus verifying the solution.

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