The solution of `(y+x+5)dy=(y-x+1)dx` is (a) `( b ) (c)log(( d ) (e) (f) (g)(( h ) (i) y+3( j ))^(( k )2( l ))( m )+( n ) (o)(( p ) (q) x+2( r ))^(( s )2( t ))( u ) (v))+( w ) (x)tan^(( y ) (z)-1( a a ))( b b ) (cc)(( d d ) y+3)/( e e )(( f f ) y+2)( g g ) (hh)+C (ii)` (jj) (kk) `( l l ) (mm)log(( n n ) (oo) (pp) (qq)(( r r ) (ss) y+3( t t ))^(( u u )2( v v ))( w w )+( x x ) (yy)(( z z ) (aaa) x-2( b b b ))^(( c c c )2( d d d ))( e e e ) (fff))+( g g g ) (hhh)tan^(( i i i ) (jjj)-1( k k k ))( l l l ) (mmm)(( n n n ) y-3)/( o o o )(( p p p ) y-2)( q q q ) (rrr)+C (sss)` (ttt) (uuu) `( v v v ) (www)log(( x x x ) (yyy) (zzz) (aaaa)(( b b b b ) (cccc) y+3( d d d d ))^(( e e e e )2( f f f f ))( g g g g )+( h h h h ) (iiii)(( j j j j ) (kkkk) x+2( l l l l ))^(( m m m m )2( n n n n ))( o o o o ) (pppp))+2( q q q q ) (rrrr)tan^(( s s s s ) (tttt)-1( u u u u ))( v v v v ) (wwww)(( x x x x ) y+3)/( y y y y )(( z z z z ) x+2)( a a a a a ) (bbbbb)+C (ccccc)` (ddddd) (eeeee) `( f f f f f ) (ggggg)log(( h h h h h ) (iiiii) (jjjjj) (kkkkk)(( l l l l l ) (mmmmm) y+3( n n n n n ))^(( o o o o o )2( p p p p p ))( q q q q q )+( r r r r r ) (sssss)(( t t t t t ) (uuuuu) x+2( v v v v v ))^(( w w w w w )2( x x x x x ))( y y y y y ) (zzzzz))-2( a a a a a a ) (bbbbbb)tan^(( c c c c c c ) (dddddd)-1( e e e e e e ))( f f f f f f ) (gggggg)(( h h h h h h ) y+3)/( i i i i i i )(( j j j j j j ) x+2)( k k k k k k ) (llllll)+C (mmmmmm)` (nnnnnn)

Options

A. `log((y+3)^(2)+(x+2)^(2))+tan^(-1)(y+3)/(y+2)+C`
B. `log((y+3)^(2)+(x-2)^(2))+tan^(-1)(y-3)/(x-2)=C`
C. `log((y+3)^(2)+(x+2)^(2))+2tan^(-1)(y+3)/(x+2)=C`
D. `log((y+3)^(2)+(x+2)^(2))-2tan^(-1)(y+3)/(x+2)=C`

Explanation

To solve the differential equation \((y + x + 5) dy = (y - x + 1) dx\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ (y + x + 5) dy - (y - x + 1) dx = 0 \]
This can be rewritten as:
\[ (y + x + 5) \frac{dy}{dx} = (y - x + 1) \]

### Step 2: Identifying Homogeneity
Notice that the equation is homogeneous. We can express \(y\) in terms of \(x\) by substituting \(y = vx\), where \(v\) is a function of \(x\). Thus, we have:
\[ dy = v dx + x \frac{dv}{dx} \]

### Step 3: Substituting into the Equation
Substituting \(y = vx\) and \(dy\) into the equation gives:
\[ (y + x + 5) \left(v dx + x \frac{dv}{dx}\right) = (vx - x + 1) dx \]
Expanding and simplifying:
\[ (vx + x + 5) \left(v dx + x \frac{dv}{dx}\right) = (vx - x + 1) dx \]

### Step 4: Simplifying the Equation
We can simplify this equation. Dividing through by \(dx\) and rearranging leads to:
\[ (v + 1 + \frac{5}{x}) \left(v + x \frac{dv}{dx}\right) = (v - 1 + \frac{1}{x}) \]

### Step 5: Separating Variables
Rearranging gives us:
\[ x \frac{dv}{dx} = \frac{(v - 1) - (v + 1 + \frac{5}{x})}{(v + 1 + \frac{5}{x})} \]
This can be further simplified to separate variables.

### Step 6: Integrating Both Sides
Integrating both sides will yield:
\[ \int \frac{1}{v^2 + 1} dv = \int \frac{1}{x} dx \]
The left-hand side integrates to \(\tan^{-1}(v)\) and the right-hand side integrates to \(\ln|x| + C\).

### Step 7: Back Substitution
Substituting back \(v = \frac{y}{x}\), we get:
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

### Step 8: Final Form
Rearranging gives us:
\[ y + 3 = k(x + 2) \]
where \(k\) is a constant derived from the integration constant.

### Conclusion
Thus, the solution to the differential equation is:
\[ \log\left(\frac{(y + 3)^2 + (x + 2)^2}{(x + 2)^2}\right) + 2\tan^{-1}\left(\frac{y + 3}{x + 2}\right) = C \]