Question 16 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=1+x2y = \sqrt{1+x^2} : y′=xy1+x2y' = \frac{xy}{1+x^2}

Correct Answer

To verify that y=1+x2y = \sqrt{1+x^2} is a solution of the differential equation y′=xy1+x2y' = \frac{xy}{1+x^2}:

Step 1: Differentiate the given function with respect to x.

y=(1+x2)1/2y = (1+x^2)^{1/2} y′=12(1+x2)−1/2⋅2x=x1+x2y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}

Step 2: Substitute y and y' into the RHS of the differential equation. RHS = xy1+x2=x⋅1+x21+x2=x1+x2\frac{xy}{1+x^2} = \frac{x \cdot \sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}

Step 3: Compare LHS and RHS. LHS = y′=x1+x2y' = \frac{x}{\sqrt{1+x^2}} and RHS = x1+x2\frac{x}{\sqrt{1+x^2}}

Since LHS = RHS, the given function is a solution of the corresponding differential equation.

Exercise: EXERCISE 9.2 | Q: 4 | (Page 7)
For More Understanding

Explanation

This is a verification problem from Exercise 9.2. The student needs to differentiate the given explicit function and substitute into the differential equation to verify equality. The differentiation uses the chain rule on the square root function.

Solution Steps

  1. Step 1: Find y' by differentiating y = (1+x2x^{2})^(1/2) using chain rule

  2. Step 2: Substitute y and y' into the RHS of the differential equation

  3. Step 3: Verify that LHS equals RHS, confirming the function is a solution

PreviousNext