Let y = `(a sin x+ (b +c) cos x ) e ^(x+d),` where a,b,c and d are parameters represent a family of curves, then differential equation for the given family of curves is given by `y'' -alpha y'+betay=0, ` then `alpha+ beta =`

Answer

To solve the problem, we need to derive the differential equation for the given family of curves represented by the function:

\[ y = (a \sin x + (b + c) \cos x) e^{(x + d)} \]

where \( a, b, c, \) and \( d \) are parameters. We will find the first and second derivatives of \( y \), and then form the differential equation in the standard form \( y'' - \alpha y' + \beta y = 0 \) to find \( \alpha + \beta \).

### Step 1: Differentiate \( y \) to find \( y' \)

Using the product rule, we differentiate \( y \):

\[ y' = \frac{d}{dx}[(a \sin x + (b + c) \cos x) e^{(x + d)}] \]

Applying the product rule:

\[ y' = (a \sin x + (b + c) \cos x) \frac{d}{dx}[e^{(x + d)}] + e^{(x + d)} \frac{d}{dx}[a \sin x + (b + c) \cos x] \]

Calculating the derivatives:

\[ \frac{d}{dx}[e^{(x + d)}] = e^{(x + d)} \]
\[ \frac{d}{dx}[a \sin x + (b + c) \cos x] = a \cos x - (b + c) \sin x \]

So, we have:

\[ y' = (a \sin x + (b + c) \cos x)e^{(x + d)} + e^{(x + d)}(a \cos x - (b + c) \sin x) \]

Combining terms:

\[ y' = e^{(x + d)} \left[(a \sin x + (b + c) \cos x) + (a \cos x - (b + c) \sin x)\right] \]

Simplifying:

\[ y' = e^{(x + d)} \left[(a \sin x - (b + c) \sin x + (b + c) \cos x + a \cos x)\right] \]

### Step 2: Differentiate \( y' \) to find \( y'' \)

Now we differentiate \( y' \):

\[ y'' = \frac{d}{dx}[y'] \]

Using the product rule again:

\[ y'' = e^{(x + d)} \left[(a \sin x + (b + c) \cos x) + (a \cos x - (b + c) \sin x)\right]' + y' \frac{d}{dx}[e^{(x + d)}] \]

Calculating the derivative of the first part:

\[ \left[(a \sin x + (b + c) \cos x) + (a \cos x - (b + c) \sin x)\right]' = a \cos x - (b + c) \sin x + a \sin x - (b + c) \cos x \]

Combining terms gives us:

\[ y'' = e^{(x + d)}\left[(a \cos x - (b + c) \sin x + a \sin x - (b + c) \cos x)\right] + y'e^{(x + d)} \]

### Step 3: Form the differential equation

Now we can express \( y'' - 2y' + 2y = 0 \):

From the previous steps, we can see that:

\[ y'' - 2y' + 2y = 0 \]

This gives us \( \alpha = 2 \) and \( \beta = 2 \).

### Step 4: Calculate \( \alpha + \beta \)

Now we sum \( \alpha \) and \( \beta \):

\[ \alpha + \beta = 2 + 2 = 4 \]

Thus, the final answer is:

\[ \alpha + \beta = 4 \]