Question 15 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=cos⁡x+Cy = \cos x + C : y′+sin⁡x=0y' + \sin x = 0

Correct Answer

To verify that the given function is a solution, we first differentiate the explicit function y=cos⁡x+Cy = \cos x + C with respect to xx. Upon differentiation, we obtain the derivative y′=−sin⁡xy' = -\sin x.

Next, we substitute this value of y′y' into the corresponding differential equation y′+sin⁡x=0y' + \sin x = 0. Calculating the L.H.S., we get (−sin⁡x)+sin⁡x=0(-\sin x) + \sin x = 0. This result is equal to the R.H.S., which is 00. Since the L.H.S. equals the R.H.S., we have verified that the given function is indeed a solution of the differential equation.

Exercise: EXERCISE 9.2 | Q: 3 | (Page 7)
For More Understanding

Explanation

The provided context outlines the standard verification method where values of derivatives are substituted into the differential equation to show that L.H.S. equals R.H.S.. Following this method, differentiating the function y=cos⁡x+Cy = \cos x + C yields y′=−sin⁡xy' = -\sin x. Substituting this into the equation y′+sin⁡x=0y' + \sin x = 0 results in an identity (0=00=0), confirming the solution as per the textbook's approach.

Solution Steps

  1. Step 1: Differentiate the function y=cos⁡x+Cy = \cos x + C with respect to xx to get y′=−sin⁡xy' = -\sin x.

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

  3. Step 3: Simplify the L.H.S.: (−sin⁡x)+sin⁡x=0(-\sin x) + \sin x = 0.

  4. Step 4: Conclude that since L.H.S. = R.H.S. = 0, the function is a solution.

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