The differential equation for `y=e^(x)(acosx+bsinx)` is

Options

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

Answer

To find the differential equation for the function \( y = e^x (a \cos x + b \sin x) \), we will differentiate the function twice and eliminate the constants \( a \) and \( b \).

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = e^x (a \cos x + b \sin x) \]

Using the product rule:
\[ \frac{dy}{dx} = e^x (a \cos x + b \sin x) + e^x \left( -a \sin x + b \cos x \right) \]

This simplifies to:
\[ \frac{dy}{dx} = e^x (a \cos x + b \sin x - a \sin x + b \cos x) \]

### Step 2: Substitute \( y \) into the derivative

We can substitute \( y \) back into the equation:
\[ \frac{dy}{dx} = y + e^x (b \cos x - a \sin x) \]

### Step 3: Differentiate again

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( y + e^x (b \cos x - a \sin x) \right) \]

Using the product rule again:
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x (b \cos x - a \sin x) + e^x \left( -b \sin x - a \cos x \right) \]

This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x (b \cos x - a \sin x - b \sin x - a \cos x) \]

### Step 4: Collect terms

Now we can combine the terms:
\[ \frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x \left( (b - a) \cos x - (a + b) \sin x \right) \]

### Step 5: Form the differential equation

Now, we can express the second derivative in terms of the first derivative and \( y \):
\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} - y = 0 \]

This is the required differential equation.

### Final Form of the Differential Equation

The final form of the differential equation is:
\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} - y = 0 \]