Question:medium

If \(a\) and \(b\) are arbitrary constants, then the differential equation representing the family of curves \[ y=a\sin(x+b) \] is

Show Hint

For \(y=a\sin(x+b)\), differentiating twice gives the relation \(y''=-y\).
  • \(\frac{d^2y}{dx^2}-y=0\)
  • \(\frac{d^2y}{dx^2}+y=0\)
  • \(\frac{d^2y}{dx^2}-y^2=0\)
  • \(\frac{dy}{dx}-y=0\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The goal is to form a differential equation by eliminating the arbitrary constants \(a\) and \(b\) from the given equation of the curve \(y = a \sin(x + b)\).
Since there are two arbitrary constants, we must differentiate the equation twice to obtain enough information to eliminate them.
Step 2: Key Formula or Approach:
Differentiate \(y\) with respect to \(x\).
Differentiate again to find the second derivative.
Substitute the original expression of \(y\) into the second derivative equation to eliminate \(a\) and \(b\).
Step 3: Detailed Explanation:

Given equation: \(y = a \sin(x + b)\) ... (1)

Differentiating equation (1) with respect to \(x\):
\[ \frac{dy}{dx} = a \cos(x + b) \cdot \frac{d}{dx}(x + b) \]
\[ \frac{dy}{dx} = a \cos(x + b) \] ... (2)

Differentiating equation (2) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = a [-\sin(x + b)] \cdot \frac{d}{dx}(x + b) \]
\[ \frac{d^2y}{dx^2} = -a \sin(x + b) \] ... (3)

Notice that the term on the right side of equation (3), \(a \sin(x + b)\), is exactly equal to \(y\) from the original equation (1).

Substitute \(y\) for \(a \sin(x + b)\) in equation (3):
\[ \frac{d^2y}{dx^2} = -y \]

Rearrange the terms to get the final differential equation:
\[ \frac{d^2y}{dx^2} + y = 0 \]

Step 4: Final Answer:
By differentiating the curve equation twice and substituting the original function back, we successfully eliminated the constants. The resulting equation is \(\frac{d^2y}{dx^2} + y = 0\).
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