The differential equation of the curve `y=e^(x) (a cos x + b sin x)` representing the given family of curves where a and b are costants , is

Options

A. `y''-2y'+2y=0`
B. `y''+2y'-2y=0`
C. `y''+2y'+2y=0`
D. None of these

Explanation

To derive the differential equation for the curve given by \( y = e^x (a \cos x + b \sin x) \), we will follow these steps:

### Step 1: Differentiate the function \( y \)
Given:
\[ y = e^x (a \cos x + b \sin x) \]
We will apply the product rule for differentiation:
\[ \frac{dy}{dx} = e^x (a \cos x + b \sin x)' + (e^x)' (a \cos x + b \sin x) \]
Calculating the derivatives:
\[ (a \cos x + b \sin x)' = -a \sin x + b \cos x \]
and
\[ (e^x)' = e^x \]
Thus, we have:
\[ \frac{dy}{dx} = e^x (-a \sin x + b \cos x) + e^x (a \cos x + b \sin x) \]
Combining the terms:
\[ \frac{dy}{dx} = e^x [(-a \sin x + b \cos x) + (a \cos x + b \sin x)] \]
This simplifies to:
\[ \frac{dy}{dx} = e^x [b \cos x + b \sin x + a \cos x - a \sin x] \]

### Step 2: Rearranging the equation
We can express this as:
\[ \frac{dy}{dx} = e^x [(b + a) \cos x + (b - a) \sin x] \]

### Step 3: Differentiate again to find the second derivative
Now we differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( e^x [(b + a) \cos x + (b - a) \sin x] \right) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = e^x [(b + a) \cos x + (b - a) \sin x]' + (e^x)' [(b + a) \cos x + (b - a) \sin x] \]
Calculating the derivative:
\[ [(b + a) \cos x + (b - a) \sin x]' = -(b + a) \sin x + (b - a) \cos x \]
Thus:
\[ \frac{d^2y}{dx^2} = e^x [-(b + a) \sin x + (b - a) \cos x] + e^x [(b + a) \cos x + (b - a) \sin x] \]
Combining these terms gives:
\[ \frac{d^2y}{dx^2} = e^x [(-b - a + b + a) \sin x + (b - a + b + a) \cos x] \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = e^x [0 \sin x + 2b \cos x] \]

### Step 4: Form the differential equation
Now we can express the relationship between \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 \]

### Final Result
Thus, the required differential equation is:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 \]

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