A function `y=f(x)` satisfies `(x+1)f^(prime)(x)-2(x^2+x)f(x)=(e^x^2)/((x+1)),AAx >-1.` If `f(0)=5,` then `f(x)` is (a) `( b ) (c)(( d ) (e) (f)(( g )3x+5)/( h )(( i ) x+1)( j ) (k) (l))( m ) e^( n ) (o) (p) x^((( q )2( r ))( s ) (t))( u ) (v)` (w) (b) `( x ) (y)(( z ) (aa) (bb)(( c c )6x+5)/( d d )(( e e ) x+1)( f f ) (gg) (hh))( i i ) e^( j j ) (kk) (ll) x^((( m m )2( n n ))( o o ) (pp))( q q ) (rr)` (ss) (c) `( d ) (e)(( f ) (g) (h)(( i )6x+5)/( j )(( k ) x+1)( l ) (m) (n))( o ) e^( p ) (q) (r) x^((( s )2( t ))( u ) (v))( w ) (x)` (y) (d) `( z ) (aa)(( b b ) (cc) (dd)(( e e )5-6x)/( f f )(( g g ) x+1)( h h ) (ii) (jj))( k k ) e^( l l ) (mm) (nn) x^((( o o )2( p p ))( q q ) (rr))( s s ) (tt)` (uu)

Options

A. `((3x+5)/(x+1))e^(x^(2))`
B. `((6x+5)/(x+1))e^(x^(2))`
C. `((6x+5)/(x+1)^(2))e^(x^(2))`
D. `((5-6x)/(x+1))e^(x^(2))`

Explanation

To solve the differential equation given by

\[ (x + 1)f'(x) - 2(x^2 + x)f(x) = \frac{e^{x^2}}{(x + 1)} \]

with the initial condition \( f(0) = 5 \), we will follow these steps:

### Step 1: Rewrite the Equation
First, we rewrite the equation in a more manageable form by dividing through by \( (x + 1) \):

\[ f'(x) - \frac{2(x^2 + x)}{(x + 1)}f(x) = \frac{e^{x^2}}{(x + 1)^2} \]

### Step 2: Simplify the Coefficient
Next, we simplify the coefficient of \( f(x) \):

\[ \frac{2(x^2 + x)}{(x + 1)} = 2x \]

Thus, the equation becomes:

\[ f'(x) - 2xf(x) = \frac{e^{x^2}}{(x + 1)^2} \]

### Step 3: Identify \( p \) and \( q \)
We can identify \( p = 2x \) and \( q = \frac{e^{x^2}}{(x + 1)^2} \).

### Step 4: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int -p \, dx} = e^{\int -2x \, dx} = e^{-x^2} \]

### Step 5: Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor:

\[ e^{-x^2}f'(x) - 2xe^{-x^2}f(x) = e^{-x^2} \cdot \frac{e^{x^2}}{(x + 1)^2} \]

This simplifies to:

\[ e^{-x^2}f'(x) - 2xe^{-x^2}f(x) = \frac{1}{(x + 1)^2} \]

### Step 6: Rewrite the Left Side
The left side can be rewritten as the derivative of a product:

\[ \frac{d}{dx}(e^{-x^2}f(x)) = \frac{1}{(x + 1)^2} \]

### Step 7: Integrate Both Sides
Integrate both sides with respect to \( x \):

\[ e^{-x^2}f(x) = \int \frac{1}{(x + 1)^2} \, dx \]

The integral on the right side is:

\[ -\frac{1}{x + 1} + C \]

Thus, we have:

\[ e^{-x^2}f(x) = -\frac{1}{x + 1} + C \]

### Step 8: Solve for \( f(x) \)
Now, multiply through by \( e^{x^2} \):

\[ f(x) = e^{x^2}\left(-\frac{1}{x + 1} + C\right) \]

### Step 9: Apply the Initial Condition
Using the initial condition \( f(0) = 5 \):

\[ 5 = e^{0}\left(-\frac{1}{0 + 1} + C\right) \implies 5 = -1 + C \implies C = 6 \]

### Step 10: Final Form of \( f(x) \)
Substituting \( C \) back into the equation gives:

\[ f(x) = e^{x^2}\left(-\frac{1}{x + 1} + 6\right) = e^{x^2}\left(\frac{6(x + 1) - 1}{x + 1}\right) = e^{x^2}\left(\frac{6x + 5}{x + 1}\right) \]

Thus, the final solution is:

\[ f(x) = \frac{6x + 5}{x + 1} e^{x^2} \]