The differential equation corresponding to the family of curves `y= e^(x) (a cos x +b sin x)`, where a and b are arbitrary constants, is

Options

A. `2y_(2)+y_(1)-3y=0`
B. `y_(2)-2y_(1)+2y=0`
C. `2y_(2)+2y_(1)-y=0`
D. `2y_(2)-y_(1)+2y=0`

Answer

To find the differential equation corresponding to the family of curves given by \( y = e^x (a \cos x + b \sin x) \), where \( a \) and \( b \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)
We start with the function:
\[ y = e^x (a \cos x + b \sin x) \]
Using the product rule for differentiation, we have:
\[ \frac{dy}{dx} = e^x (a \cos x + b \sin x) + e^x \left( -a \sin x + b \cos x \right) \]
This simplifies to:
\[ \frac{dy}{dx} = e^x \left( (a \cos x + b \sin x) + (-a \sin x + b \cos x) \right) \]
\[ \frac{dy}{dx} = e^x \left( a \cos x + b \sin x - a \sin x + b \cos x \right) \]
\[ \frac{dy}{dx} = e^x \left( (a + b) \cos x + (b - a) \sin x \right) \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( e^x \left( (a + b) \cos x + (b - a) \sin x \right) \right) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = e^x \left( (a + b) \cos x + (b - a) \sin x \right) + e^x \left( -(a + b) \sin x + (b - a) \cos x \right) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = e^x \left( (a + b) \cos x + (b - a) \sin x - (a + b) \sin x + (b - a) \cos x \right) \]
\[ \frac{d^2y}{dx^2} = e^x \left( ((a + b) + (b - a)) \cos x + ((b - a) - (a + b)) \sin x \right) \]
\[ \frac{d^2y}{dx^2} = e^x \left( 2b \cos x - 2a \sin x \right) \]

### Step 3: Formulate the differential equation
From the expressions for \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \), we can express the second derivative in terms of \( y \) and \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 \]

### Final Answer
Thus, the differential equation corresponding to the family of curves is:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 \]