If `x(t)` is a solution of `((1+t)dy)/(dx)-t y=1` and `y(0)=-1` then `y(1)` is (a) `( b ) (c)-( d )1/( e )2( f ) (g) (h)` (i) (b) `( j ) (k) e+( l )1/( m )2( n ) (o) (p)` (q) (c) `( d ) (e) e-( f )1/( g )2( h ) (i) (j)` (k) (d) `( l ) (m) (n)1/( o )2( p ) (q) (r)` (s)

Options

A. `-(1)/(2)`
B. `e+(1)/(2)`
C. `e-(1)/(2)`
D. `(1)/(2)`

Explanation

To solve the differential equation given by

\[ (1+t) \frac{dy}{dx} - ty = 1 \]

with the initial condition \( y(0) = -1 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We can rewrite the equation as:

\[ \frac{dy}{dx} - \frac{t}{1+t} y = \frac{1}{1+t} \]

This is now in the standard form of a linear first-order differential equation:

\[ \frac{dy}{dx} + P(t)y = Q(t) \]

where \( P(t) = -\frac{t}{1+t} \) and \( Q(t) = \frac{1}{1+t} \).

### Step 2: Find the integrating factor
The integrating factor \( \mu(t) \) is given by:

\[ \mu(t) = e^{\int P(t) dt} = e^{\int -\frac{t}{1+t} dt} \]

To compute this integral, we can use the substitution \( u = 1 + t \), which gives \( du = dt \) and \( t = u - 1 \). Thus, the integral becomes:

\[ \int -\frac{t}{1+t} dt = \int -\frac{u-1}{u} du = \int -1 + \frac{1}{u} du = -u + \ln|u| + C = -(1+t) + \ln(1+t) \]

So, the integrating factor is:

\[ \mu(t) = e^{-(1+t) + \ln(1+t)} = \frac{1+t}{e^{1+t}} \]

### Step 3: Multiply through by the integrating factor
Now we multiply the entire differential equation by the integrating factor:

\[ \frac{1+t}{e^{1+t}} \frac{dy}{dx} - \frac{t(1+t)}{(1+t)e^{1+t}} y = \frac{(1+t)}{(1+t)e^{1+t}} \]

This simplifies to:

\[ \frac{1+t}{e^{1+t}} \frac{dy}{dx} - \frac{t}{e^{1+t}} y = \frac{1}{e^{1+t}} \]

### Step 4: Solve the equation
The left-hand side can be rewritten as:

\[ \frac{d}{dx} \left( y \frac{1+t}{e^{1+t}} \right) = \frac{1}{e^{1+t}} \]

Integrating both sides with respect to \( t \):

\[ y \frac{1+t}{e^{1+t}} = \int \frac{1}{e^{1+t}} dt \]

The integral on the right can be computed as:

\[ \int e^{-(1+t)} dt = -e^{-(1+t)} + C \]

Thus, we have:

\[ y \frac{1+t}{e^{1+t}} = -e^{-(1+t)} + C \]

### Step 5: Solve for \( y \)
Now, solving for \( y \):

\[ y = \frac{-e^{-(1+t)} + C}{\frac{1+t}{e^{1+t}}} = \frac{-e^{-(1+t)} + C}{\frac{1+t}{e^{1+t}}} \cdot \frac{e^{1+t}}{1+t} \]

This simplifies to:

\[ y = -1 + C \cdot e^{1+t} \]

### Step 6: Apply the initial condition
Using the initial condition \( y(0) = -1 \):

\[ -1 = -1 + C \cdot e^{1} \]

This implies \( C = 0 \). Therefore, the solution simplifies to:

\[ y = -1 \]

### Step 7: Find \( y(1) \)
Finally, we need to find \( y(1) \):

\[ y(1) = -1 \]

### Conclusion
Thus, the final answer is:

\[ \boxed{-1} \]