Question 88 of 96intermediate🔧 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⁡y=xy - \cos y = x : (ysin⁡y+cos⁡y+x)y′=y(y \sin y + \cos y + x) y' = y

Correct Answer

To verify the solution, we first differentiate the implicit function y−cos⁡y=xy - \cos y = x with respect to xx. Using the chain rule, we obtain dydx−(−sin⁡ydydx)=1\frac{dy}{dx} - (-\sin y \frac{dy}{dx}) = 1, which simplifies to y′(1+sin⁡y)=1y'(1 + \sin y) = 1 or y′=11+sin⁡yy' = \frac{1}{1 + \sin y}.

Next, we substitute x=y−cos⁡yx = y - \cos y from the function into the differential equation (ysin⁡y+cos⁡y+x)y′=y(y \sin y + \cos y + x) y' = y. The L.H.S. becomes (ysin⁡y+cos⁡y+y−cos⁡y)y′(y \sin y + \cos y + y - \cos y) y', which simplifies to y(1+sin⁡y)y′y(1 + \sin y) y'. Substituting the derived value of y′y', we get y(1+sin⁡y)⋅11+sin⁡y=yy(1 + \sin y) \cdot \frac{1}{1 + \sin y} = y.

Since L.H.S. = R.H.S., the given function is a solution of the corresponding differential equation.

Exercise: EXERCISE 9.2 | Q: 8 | (Chapter: 7)
For More Understanding

Explanation

The solution follows the standard verification method illustrated in the example preceding Exercise 9.2 in the provided context. It involves differentiating the given function to find the derivative, substituting the function and its derivative into the differential equation, and simplifying to show that the Left Hand Side equals the Right Hand Side.

Solution Steps

  1. Step 1: Differentiate the given function y−cos⁡y=xy - \cos y = x with respect to xx to find y′y'.

  2. Step 2: Substitute the expression for xx (from the function) and y′y' into the differential equation.

  3. Step 3: Simplify the equation to verify L.H.S. = R.H.S.

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