A continuously differentiable function `phi(x)` in `(0,pi)` satisfying `y'=1+y^(2),y(0)=0=y(pi)` is

Options

A. tan x
B. `x(x-pi)`
C. `(x-pi) (1-e^(x))`
D. Not possible

Answer

To solve the given problem, we need to find a continuously differentiable function \( \phi(x) \) in the interval \( (0, \pi) \) that satisfies the differential equation:

\[ \frac{dy}{dx} = 1 + y^2 \]

with the boundary conditions \( y(0) = 0 \) and \( y(\pi) = 0 \).

### Step 1: Separate the Variables
We start by separating the variables in the differential equation:

\[ \frac{dy}{1 + y^2} = dx \]

**Hint:** Look for a way to isolate \( y \) on one side and \( x \) on the other.

### Step 2: Integrate Both Sides
Next, we integrate both sides:

\[ \int \frac{dy}{1 + y^2} = \int dx \]

The left-hand side integrates to \( \tan^{-1}(y) \), and the right-hand side integrates to \( x + C \):

\[ \tan^{-1}(y) = x + C \]

**Hint:** Remember the integral of \( \frac{1}{1+y^2} \) is \( \tan^{-1}(y) \).

### Step 3: Solve for \( y \)
Now, we solve for \( y \):

\[ y = \tan(x + C) \]

**Hint:** Keep in mind that \( y \) is expressed in terms of the tangent function.

### Step 4: Apply the Boundary Conditions
We have two conditions to apply: \( y(0) = 0 \) and \( y(\pi) = 0 \).

1. For \( y(0) = 0 \):

\[ 0 = \tan(0 + C) \implies C = 0 \]

2. For \( y(\pi) = 0 \):

\[ 0 = \tan(\pi + C) \implies C = -\pi \]

**Hint:** Use the properties of the tangent function to find the constants.

### Step 5: Substitute Back the Value of \( C \)
Now substituting \( C = -\pi \) back into our equation for \( y \):

\[ y = \tan(x - \pi) \]

Using the property of tangent:

\[ y = -\tan(x) \]

**Hint:** Remember that \( \tan(x - \pi) = -\tan(x) \).

### Step 6: Check Continuity
The function \( y = -\tan(x) \) is not continuous in the interval \( (0, \pi) \) because \( \tan(x) \) has vertical asymptotes at \( x = \frac{\pi}{2} \).

**Hint:** Analyze the behavior of the tangent function to determine continuity.

### Conclusion
Since \( y = -\tan(x) \) is not a continuously differentiable function in the interval \( (0, \pi) \), we conclude that no such function \( \phi(x) \) exists that satisfies the given conditions.

**Final Answer:** The continuously differentiable function \( \phi(x) \) does not exist in the interval \( (0, \pi) \).