Question 17 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=Axy = Ax : xy′=y(x≠0)xy' = y (x \neq 0)

Correct Answer

We are given the function y = Ax, where A is a constant, and the differential equation xy' = y with x ≠\neq 0. To verify, we first differentiate the function with respect to x.

Differentiating y = Ax gives y' = A, since A is constant. Now substitute y and y' into the differential equation. The left-hand side becomes x y' = x · A = A x. The right-hand side is y = A x. Thus, LHS = RHS for all x ≠\neq 0.

Therefore, y = Ax satisfies the given differential equation, and it is indeed a solution.

Exercise: EXERCISE 9.2 | Q: 5 | (Page 7)
For More Understanding

Explanation

The verification follows the standard method shown in the textbook: compute the derivative of the given function and substitute both the function and its derivative into the differential equation. If both sides match, the function is a solution. Here, the derivative is constant, and substitution yields identical expressions, confirming the solution.

Solution Steps

  1. Step 1: Differentiate y = Ax to get y' = A.

  2. Step 2: Substitute into xy' = y: LHS = x·A = Ax, RHS = Ax.

  3. Step 3: Since LHS = RHS for all x ≠\neq 0, the function is a solution.

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