The differential equation whose general solution is given by `y=(c_1cos(x+c_2)-(c_3e^((-x+c4))+(c_5sinx),` where `c_1,c_2,c_3,c_4,c_5` are arbitrary constants, is (a) `( b ) (c) (d)(( e ) (f) d^(( g )4( h ))( i ) y)/( j )(( k ) d (l) x^(( m )4( n ))( o ))( p ) (q)-( r )(( s ) (t) d^(( u )2( v ))( w ) y)/( x )(( y ) d (z) x^(( a a )2( b b ))( c c ))( d d ) (ee)+y=0( f f )` (gg) (hh) `( i i ) (jj) (kk)(( l l ) (mm) d^(( n n )3( o o ))( p p ) y)/( q q )(( r r ) d (ss) x^(( t t )3( u u ))( v v ))( w w ) (xx)+( y y )(( z z ) (aaa) d^(( b b b )2( c c c ))( d d d ) y)/( e e e )(( f f f ) d (ggg) x^(( h h h )2( i i i ))( j j j ))( k k k ) (lll)+( m m m )(( n n n ) dy)/( o o o )(( p p p ) dx)( q q q ) (rrr)+y=0( s s s )` (ttt) (uuu) `( v v v ) (www) (xxx)(( y y y ) (zzz) d^(( a a a a )5( b b b b ))( c c c c ))/( d d d d )(( e e e e ) d (ffff) x^(( g g g g )5( h h h h ))( i i i i ))( j j j j ) (kkkk)+y=0( l l l l )` (mmmm) (nnnn) `( o o o o ) (pppp) (qqqq)(( r r r r ) (ssss) d^(( t t t t )3( u u u u ))( v v v v ) y)/( w w w w )(( x x x x ) d (yyyy) x^(( z z z z )3( a a a a a ))( b b b b b ))( c c c c c ) (ddddd)-( e e e e e )(( f f f f f ) (ggggg) d^(( h h h h h )2( i i i i i ))( j j j j j ) y)/( k k k k k )(( l l l l l ) d (mmmmm) x^(( n n n n n )2( o o o o o ))( p p p p p ))( q q q q q ) (rrrrr)+( s s s s s )(( t t t t t ) dy)/( u u u u u )(( v v v v v ) dx)( w w w w w ) (xxxxx)-y=0( y y y y y )` (zzzzz)

Options

A. `(d^(4)y)/(dx^(4)) - (d^(2)y)/(dx^(2))+y=0`
B. `(d^(3)y)/(dx^(3))+(d^(2)y)/(dx^(2))+(dy)/(dx)+y=0`
C. `(d^(5)y)/(dx^(5))+y=0`
D. `(d^(3)y)/(dx^(3))-(d^(2)y)/(dx^(2))+(dy)/(dx)-y=0`

Explanation

To find the differential equation whose general solution is given by

\[ y = c_1 \cos(x + c_2) - (c_3 e^{-x + c_4}) + c_5 \sin x \]

where \(c_1, c_2, c_3, c_4, c_5\) are arbitrary constants, we will follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten as:

\[ y = c_1 \cos x \cos c_2 - c_1 \sin x \sin c_2 - c_3 e^{-x + c_4} + c_5 \sin x \]

### Step 2: Differentiate the equation
We differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = -c_1 \sin x \cos c_2 + c_1 \cos x \sin c_2 + c_3 e^{-x + c_4} - c_5 \cos x \]

### Step 3: Differentiate again
Now we differentiate \(\frac{dy}{dx}\) to get the second derivative:

\[ \frac{d^2y}{dx^2} = -c_1 \cos x \cos c_2 - c_1 \sin x \sin c_2 - c_3 e^{-x + c_4} + c_5 \sin x \]

### Step 4: Differentiate a third time
Next, we differentiate \(\frac{d^2y}{dx^2}\) to find the third derivative:

\[ \frac{d^3y}{dx^3} = c_1 \sin x \cos c_2 - c_1 \cos x \sin c_2 + c_3 e^{-x + c_4} + c_5 \cos x \]

### Step 5: Combine the derivatives
Now, we will combine the equations derived from the first, second, and third derivatives.

Adding the first and third derivatives:

\[ \frac{d^2y}{dx^2} + y = -2c_3 e^{-x + c_4} \]

Adding the second and fourth derivatives:

\[ \frac{d^3y}{dx^3} + \frac{dy}{dx} = 2c_3 e^{-x + c_4} \]

### Step 6: Final equation
Combining these results, we arrive at the final differential equation:

\[ \frac{d^3y}{dx^3} + \frac{dy}{dx} + y = 0 \]

Thus, the differential equation whose general solution is given by the original equation is:

\[ \frac{d^3y}{dx^3} + \frac{dy}{dx} + y = 0 \]