The differential equation whose solution is y=A sin x + B cos x , is

Options

A. `(d^2y)/(dx^2)+ y =0`
B. `(d^2y)/(dx^2)-y=0`
C. `(dy)/(dx)+ y =0`
D. `(dy)/(dx)-y =0`

Answer

To find the differential equation whose solution is given by \( y = A \sin x + B \cos x \), we will follow these steps:

### Step 1: Differentiate the given function
We start with the function:
\[ y = A \sin x + B \cos x \]
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = A \cos x - B \sin x \]

### Step 2: Differentiate again to find the second derivative
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative:
\[ \frac{d^2y}{dx^2} = -A \sin x - B \cos x \]

### Step 3: Relate the second derivative to the original function
Notice that the second derivative can be expressed in terms of \( y \):
\[ \frac{d^2y}{dx^2} = - (A \sin x + B \cos x) = -y \]

### Step 4: Formulate the differential equation
From the relationship we found in Step 3, we can write the differential equation as:
\[ \frac{d^2y}{dx^2} + y = 0 \]

### Final Answer
Thus, the differential equation whose solution is \( y = A \sin x + B \cos x \) is:
\[ \frac{d^2y}{dx^2} + y = 0 \]

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