The differential equation of the family of curves `y=e^x(Acosx+Bsinx),` where `A` and `B` are arbitrary constants is (a) `( b ) (c) (d)(( e ) (f) d^(( g )2( h ))( i ) y)/( j )(( k ) d (l) x^(( m )2( n ))( o ))( p ) (q)-2( r )(( s ) dy)/( t )(( u ) dx)( v ) (w)+2y=0( x )` (y) (z) `( a a ) (bb) (cc)(( d d ) (ee) d^(( f f )2( g g ))( h h ) y)/( i i )(( j j ) d (kk) x^(( l l )2( m m ))( n n ))( o o ) (pp)+2( q q )(( r r ) dy)/( s s )(( t t ) dx)( u u ) (vv)+2y=0( w w )` (xx) (yy) `( z z ) (aaa) (bbb)(( c c c ) (ddd) d^(( e e e )2( f f f ))( g g g ) y)/( h h h )(( i i i ) d (jjj) x^(( k k k )2( l l l ))( m m m ))( n n n ) (ooo)+( p p p ) (qqq)(( r r r ) (sss) (ttt)(( u u u ) dy)/( v v v )(( w w w ) dx)( x x x ) (yyy) (zzz))^(( a a a a )2( b b b b ))( c c c c )+y=0( d d d d )` (eeee) (ffff) `( g g g g ) (hhhh) (iiii)(( j j j j ) (kkkk) d^(( l l l l )2( m m m m ))( n n n n ) y)/( o o o o )(( p p p p ) d (qqqq) x^(( r r r r )2( s s s s ))( t t t t ))( u u u u ) (vvvv)-7( w w w w )(( x x x x ) dy)/( y y y y )(( z z z z ) dx)( a a a a a ) (bbbbb)+2y=0( c c c c c )` (ddddd)

Options

A. `(d^2y)/(dx^2)+2(dy)/(dx)+2y=0`
B. `(d^2y)/(dx^2)-2 (dy)/(dx)+2y=0`
C. `(d^2y)/(dx^2)-2 (dy )/(dx)-2y=0`
D. none of these

Answer

To find the differential equation of 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 \( x \)
Given:
\[ y = e^x (A \cos x + B \sin x) \]
We will differentiate \( y \) with respect to \( x \) using the product rule:
\[ \frac{dy}{dx} = e^x (A \cos x + B \sin x)' + (e^x)' (A \cos x + B \sin x) \]
Calculating the derivatives:
\[ (A \cos x + B \sin x)' = -A \sin x + B \cos x \]
So,
\[ \frac{dy}{dx} = e^x (-A \sin x + B \cos x) + e^x (A \cos x + B \sin x) \]
This simplifies to:
\[ \frac{dy}{dx} = e^x (B \cos x - A \sin x + A \cos x + B \sin x) = e^x ((B + A) \cos x + (B - A) \sin x) \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( e^x ((B + A) \cos x + (B - A) \sin x) \right) \]
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:
\[ ((B + A) \cos x + (B - A) \sin x)' = -(B + A) \sin x + (B - A) \cos x \]
So,
\[ \frac{d^2y}{dx^2} = e^x (-(B + A) \sin x + (B - A) \cos x) + e^x ((B + A) \cos x + (B - A) \sin x) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = e^x \left( (B - A) \cos x - (B + A) \sin x + (B + A) \cos x + (B - A) \sin x \right) \]
Combining like terms:
\[ \frac{d^2y}{dx^2} = e^x \left( 2B \cos x + 2A \sin x \right) \]

### Step 3: Substitute back to eliminate \( A \) and \( B \)
From the original equation, we can express \( A \) and \( B \) in terms of \( y \) and its derivatives. We have:
\[ y = e^x (A \cos x + B \sin x) \]
This gives us:
\[ A \cos x + B \sin x = \frac{y}{e^x} \]
Using the expressions for \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \), we can eliminate \( A \) and \( B \) to form a differential equation.

### Step 4: Form the differential equation
We can rearrange the terms to get:
\[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 \]

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

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