The solution of the differential equation `{1+xsqrt((x^2+y^2))}dx+{sqrt((x^2+y^2))-1}ydy=0` is equal to (a) `( b ) (c) (d) x^(( e )2( f ))( g )+( h )(( i ) (j) y^(( k )2( l ))( m ))/( n )2( o ) (p)+( q )1/( r )3( s ) (t) (u) (v)(( w ) (x) (y) x^(( z )2( a a ))( b b )+( c c ) y^(( d d )2( e e ))( f f ) (gg))^(( h h ) (ii) (jj)3/( k k )2( l l ) (mm) (nn))( o o )=c (pp)` (qq) (rr) `( s s ) (tt) x+( u u )(( v v ) (ww) y^(( x x )3( y y ))( z z ))/( a a a )3( b b b ) (ccc)+( d d d )1/( e e e )2( f f f ) (ggg) (hhh) (iii)(( j j j ) (kkk) (lll) x^(( m m m )2( n n n ))( o o o )+( p p p ) y^(( q q q )2( r r r ))( s s s ) (ttt))^(( u u u ) (vvv) (www)1/( x x x )2( y y y ) (zzz) (aaaa))( b b b b )=c (cccc)` (dddd) (eeee) `( f f f f ) (gggg) x-( h h h h )(( i i i i ) (jjjj) y^(( k k k k )2( l l l l ))( m m m m ))/( n n n n )2( o o o o ) (pppp)+( q q q q )1/( r r r r )3( s s s s ) (tttt) (uuuu) (vvvv)(( w w w w ) (xxxx) (yyyy) x^(( z z z z )2( a a a a a ))( b b b b b )+( c c c c c ) y^(( d d d d d )2( e e e e e ))( f f f f f ) (ggggg))^(( h h h h h ) (iiiii) (jjjjj)3/( k k k k k )2( l l l l l ) (mmmmm) (nnnnn))( o o o o o )=c (ppppp)` (qqqqq)

Options

A. `x^(2)+y^(2)/2+1/3(x^(2)+y^(2))^(3//2)=c`
B. `x-y^(3)/3+1/2(x^(2)+y^(2))^(1//2)=c`
C. `x-y^(2)/2+1/3(x^(2)+y^(2))^(3//2)=c`
D. None of these

Explanation

To solve the differential equation

\[ (1 + x\sqrt{x^2 + y^2})dx + (\sqrt{x^2 + y^2} - 1)ydy = 0, \]

we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the terms in the differential equation:

\[ dx + x\sqrt{x^2 + y^2}dx + (\sqrt{x^2 + y^2} - 1)ydy = 0. \]

This can be rewritten as:

\[ dx - y dy + x\sqrt{x^2 + y^2}dx + y\sqrt{x^2 + y^2}dy = 0. \]

### Step 2: Factoring Out Common Terms
Next, we factor out \(\sqrt{x^2 + y^2}\):

\[ dx - y dy + \sqrt{x^2 + y^2}(x dx + y dy) = 0. \]

### Step 3: Identifying a Potential Function
We can rewrite the equation in a more manageable form:

\[ dx - y dy + \sqrt{x^2 + y^2}(x dx + y dy) = 0. \]

This suggests that we can find a potential function \(F(x, y)\) such that:

\[ \frac{\partial F}{\partial x} = 1 + x\sqrt{x^2 + y^2}, \]
\[ \frac{\partial F}{\partial y} = \sqrt{x^2 + y^2} - 1. \]

### Step 4: Integrating with Respect to \(x\)
Integrating \(\frac{\partial F}{\partial x}\) with respect to \(x\):

\[ F(x, y) = \int (1 + x\sqrt{x^2 + y^2})dx. \]

This gives:

\[ F(x, y) = x + \frac{1}{2}(x^2 + y^2)^{3/2} + g(y), \]

where \(g(y)\) is a function of \(y\) alone.

### Step 5: Integrating with Respect to \(y\)
Next, we differentiate \(F(x, y)\) with respect to \(y\) and set it equal to \(\frac{\partial F}{\partial y}\):

\[ \frac{\partial F}{\partial y} = \frac{1}{2} \cdot 3(x^2 + y^2)^{1/2} \cdot 2y = 3y(x^2 + y^2)^{1/2}. \]

Setting this equal to \(\sqrt{x^2 + y^2} - 1\), we can solve for \(g(y)\).

### Step 6: Finalizing the Function
After integrating and simplifying, we find:

\[ F(x, y) = x - \frac{y^2}{2} + \frac{1}{3}(x^2 + y^2)^{3/2} = c, \]

where \(c\) is a constant.

### Step 7: Conclusion
Thus, the solution to the differential equation is:

\[ x - \frac{y^2}{2} + \frac{1}{3}(x^2 + y^2)^{3/2} = c. \]