The differential equation whose solution is `y=c_1 cos ax + c_2 sin ax` is

Options

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

Answer

To find the differential equation whose solution is given by \( y = c_1 \cos(ax) + c_2 \sin(ax) \), we will follow these steps:

### Step 1: Differentiate the given function
We start with the function:
\[ y = c_1 \cos(ax) + c_2 \sin(ax) \]
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(c_1 \cos(ax)) + \frac{d}{dx}(c_2 \sin(ax)) \]
Using the chain rule, the derivatives are:
\[ \frac{dy}{dx} = c_1 (-a \sin(ax)) + c_2 (a \cos(ax)) = -ac_1 \sin(ax) + ac_2 \cos(ax) \]

### Step 2: Differentiate again to find the second derivative
Next, we differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-ac_1 \sin(ax) + ac_2 \cos(ax)) \]
Applying the chain rule again:
\[ \frac{d^2y}{dx^2} = -ac_1 (a \cos(ax)) - ac_2 (a \sin(ax)) = -a^2 c_1 \cos(ax) - a^2 c_2 \sin(ax) \]

### Step 3: Express the second derivative in terms of \( y \)
We can factor out \(-a^2\) from the expression:
\[ \frac{d^2y}{dx^2} = -a^2 (c_1 \cos(ax) + c_2 \sin(ax)) \]
Since \( y = c_1 \cos(ax) + c_2 \sin(ax) \), we can substitute \( y \) into the equation:
\[ \frac{d^2y}{dx^2} = -a^2 y \]

### Step 4: Rearranging the equation
Rearranging gives us the standard form of the second-order linear differential equation:
\[ \frac{d^2y}{dx^2} + a^2 y = 0 \]

### Conclusion
Thus, the differential equation whose solution is \( y = c_1 \cos(ax) + c_2 \sin(ax) \) is:
\[ \frac{d^2y}{dx^2} + a^2 y = 0 \]