If y(t) is solution of `(t+1)(dy)/(dt) -ty =1, y(0)= -1`. At t = 1 the solution is

Options

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

Answer

To solve the differential equation \((t+1)\frac{dy}{dt} - ty = 1\) with the initial condition \(y(0) = -1\) and find the value of \(y\) at \(t = 1\), we can follow these steps:

### Step 1: Rewrite the differential equation
We start with the given equation:
\[ (t+1)\frac{dy}{dt} - ty = 1 \]
We can rearrange it to isolate \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} - \frac{t}{t+1}y = \frac{1}{t+1} \]

### Step 2: Identify \(p(t)\) and \(q(t)\)
In the standard form \(\frac{dy}{dt} + p(t)y = q(t)\), we identify:
- \(p(t) = -\frac{t}{t+1}\)
- \(q(t) = \frac{1}{t+1}\)

### Step 3: Find the integrating factor
The integrating factor \(\mu(t)\) is given by:
\[ \mu(t) = e^{\int p(t) dt} = e^{\int -\frac{t}{t+1} dt} \]
To compute the integral, we can simplify:
\[ \int -\frac{t}{t+1} dt = -\int \left(1 - \frac{1}{t+1}\right) dt = -\left(t - \log(t+1)\right) + C \]
Thus,
\[ \mu(t) = e^{-t + \log(t+1)} = \frac{t+1}{e^t} \]

### Step 4: Multiply through by the integrating factor
We multiply the entire differential equation by the integrating factor:
\[ \frac{(t+1)}{e^t} \frac{dy}{dt} - \frac{t(t+1)}{(t+1)e^t}y = \frac{(t+1)}{(t+1)e^t} \]
This simplifies to:
\[ \frac{(t+1)}{e^t} \frac{dy}{dt} - \frac{t}{e^t}y = \frac{1}{e^t} \]

### Step 5: Integrate both sides
The left side can be expressed as the derivative of a product:
\[ \frac{d}{dt}\left(\frac{(t+1)y}{e^t}\right) = \frac{1}{e^t} \]
Integrating both sides gives:
\[ \frac{(t+1)y}{e^t} = \int \frac{1}{e^t} dt + C \]
The integral evaluates to:
\[ \frac{(t+1)y}{e^t} = -e^{-t} + C \]

### Step 6: Solve for \(y\)
Multiplying through by \(e^t\):
\[ (t+1)y = -1 + Ce^t \]
Thus,
\[ y = \frac{-1 + Ce^t}{t+1} \]

### Step 7: Apply the initial condition
Using the initial condition \(y(0) = -1\):
\[ -1 = \frac{-1 + C}{1} \]
This simplifies to:
\[ -1 = -1 + C \implies C = 0 \]
So, the solution simplifies to:
\[ y = \frac{-1}{t+1} \]

### Step 8: Find \(y\) at \(t = 1\)
Now we substitute \(t = 1\):
\[ y(1) = \frac{-1}{1+1} = \frac{-1}{2} \]

### Final Answer
Thus, the value of \(y\) at \(t = 1\) is:
\[ \boxed{-\frac{1}{2}} \]