Question 94 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: x+y=tan⁡−1yx + y = \tan^{-1}y : y2y′+y2+1=0y^2 y' + y^2 + 1 = 0

Correct Answer

To verify that x+y=tan⁡−1yx + y = \tan^{-1}y is a solution of the differential equation y2y′+y2+1=0y^2 y' + y^2 + 1 = 0, we differentiate the given implicit function with respect to xx.

Differentiating both sides: ddx(x+y)=ddx(tan⁡−1y)\frac{d}{dx}(x + y) = \frac{d}{dx}(\tan^{-1}y) This gives us: 1+dydx=11+y2⋅dydx1 + \frac{dy}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx}

Let y′=dydxy' = \frac{dy}{dx}. Rearranging: 1+y′=y′1+y21 + y' = \frac{y'}{1 + y^2} Multiplying by (1+y2)(1 + y^2): (1+y′)(1+y2)=y′(1 + y')(1 + y^2) = y' Expanding: 1+y2+y′+y2y′=y′1 + y^2 + y' + y^2 y' = y' Simplifying: 1+y2+y2y′=01 + y^2 + y^2 y' = 0

Therefore, y2y′+y2+1=0y^2 y' + y^2 + 1 = 0, which is exactly the given differential equation. Hence, the given function is verified as a solution.

Exercise: EXERCISE 9.2 | Q: 9 | (Chapter: 7)
For More Understanding

Explanation

This is Exercise 9.2, Question 9 from NCERT Class 12 Mathematics. The question asks to verify that the implicit function x + y = arctan(y) satisfies the differential equation y²y' + y² + 1 = 0. The solution involves implicit differentiation - differentiating both sides with respect to x, then algebraically manipulating to arrive at the given differential equation.

Solution Steps

  1. Step 1: Differentiate the implicit function x + y = tan⁻¹y with respect to x

  2. Step 2: Apply chain rule: 1 + y' = (1/(1+y²)) · y'

  3. Step 3: Multiply both sides by (1 + y²) and expand

  4. Step 4: Simplify to get y²y' + y² + 1 = 0

  5. Step 5: Conclude that the LHS matches the given differential equation

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