What is the only solution of the initial value problem `y ' = t (1 + y) , y (0) = 0 `?

Options

A. A) `y = -1 + e^(t^(2) //2)`
B. B) `y = 1 + e^(t^(2)//2)`
C. C) `y = -t`
D. D) `y = t`

Answer

To solve the initial value problem \( y' = t(1 + y) \) with the initial condition \( y(0) = 0 \), we will follow these steps:

### Step 1: Rewrite the differential equation
We start with the given differential equation:
\[ y' = t(1 + y) \]
This can be rewritten as:
\[ \frac{dy}{dt} = t(1 + y) \]

### Step 2: Separate the variables
We can separate the variables \( y \) and \( t \):
\[ \frac{dy}{1 + y} = t \, dt \]

### Step 3: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dy}{1 + y} = \int t \, dt \]
The left side integrates to:
\[ \log |1 + y| \]
The right side integrates to:
\[ \frac{t^2}{2} + C \]
Thus, we have:
\[ \log |1 + y| = \frac{t^2}{2} + C \]

### Step 4: Solve for \( y \)
To solve for \( y \), we exponentiate both sides:
\[ |1 + y| = e^{\frac{t^2}{2} + C} = e^C e^{\frac{t^2}{2}} \]
Let \( K = e^C \), then:
\[ 1 + y = K e^{\frac{t^2}{2}} \]
This gives us:
\[ y = K e^{\frac{t^2}{2}} - 1 \]

### Step 5: Apply the initial condition
We apply the initial condition \( y(0) = 0 \):
\[ 0 = K e^{\frac{0^2}{2}} - 1 \]
This simplifies to:
\[ 0 = K - 1 \implies K = 1 \]

### Step 6: Write the final solution
Substituting \( K \) back into the equation for \( y \):
\[ y = 1 e^{\frac{t^2}{2}} - 1 = e^{\frac{t^2}{2}} - 1 \]

Thus, the only solution of the initial value problem is:
\[ y = e^{\frac{t^2}{2}} - 1 \]