Question 18 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=xsin⁡xy = x \sin x : xy′=y+xx2−y2(x≠0 and x>y or x<−y)xy' = y + x \sqrt{x^2 - y^2} (x \neq 0 \text{ and } x > y \text{ or } x < -y)

Correct Answer

To verify the solution, we first differentiate the given function y=xsin⁡xy = x \sin x with respect to xx. Using the product rule, we find the derivative y′=xcos⁡x+sin⁡xy' = x \cos x + \sin x. Substituting this into the LHS of the differential equation, we get xy′=x(xcos⁡x+sin⁡x)=x2cos⁡x+xsin⁡xxy' = x(x \cos x + \sin x) = x^2 \cos x + x \sin x.

Next, we substitute y=xsin⁡xy = x \sin x into the RHS of the equation: y+xx2−y2y + x \sqrt{x^2 - y^2}. This simplifies to xsin⁡x+xx2−(xsin⁡x)2=xsin⁡x+xx2−x2sin⁡2xx \sin x + x \sqrt{x^2 - (x \sin x)^2} = x \sin x + x \sqrt{x^2 - x^2 \sin^2 x}. Using the identity 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x, the expression becomes xsin⁡x+xx2cos⁡2x=xsin⁡x+x(xcos⁡x)=xsin⁡x+x2cos⁡xx \sin x + x \sqrt{x^2 \cos^2 x} = x \sin x + x(x \cos x) = x \sin x + x^2 \cos x.

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

Exercise: EXERCISE 9.2 | Q: 6 | (Page 7)
For More Understanding

Explanation

The solution follows the standard verification method illustrated in the provided context for similar exercises (like verifying solutions in Exercise 9.2). The steps involve differentiating the explicit function y=xsin⁡xy = x \sin x and substituting both yy and y′y' into the differential equation to prove equality between the Left Hand Side (LHS) and Right Hand Side (RHS).

Solution Steps

  1. Step 1: Differentiate the function y=xsin⁡xy = x \sin x with respect to xx to get y′=xcos⁡x+sin⁡xy' = x \cos x + \sin x.

  2. Step 2: Calculate the LHS by substituting y′y' into xy′xy': x(xcos⁡x+sin⁡x)=x2cos⁡x+xsin⁡xx(x \cos x + \sin x) = x^2 \cos x + x \sin x.

  3. Step 3: Calculate the RHS by substituting yy into y+xx2−y2y + x \sqrt{x^2 - y^2}: xsin⁡x+xx2−x2sin⁡2x=xsin⁡x+x2cos⁡xx \sin x + x \sqrt{x^2 - x^2 \sin^2 x} = x \sin x + x^2 \cos x.

  4. Step 4: Compare LHS and RHS to show they are equal, verifying the solution.

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