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

Options

A. ` Y= tan x `
B. ` y= x(x- pi )`
C. `y= (x- pi ) ( 1 - e^(x))`
D. not possible

Answer

To solve the differential equation \( y' = 1 + y^2 \) with the initial condition \( y(0) = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation can be rewritten as:
\[ \frac{dy}{dx} = 1 + y^2 \]

### Step 2: Separate Variables
We can separate the variables \( y \) and \( x \):
\[ \frac{dy}{1 + y^2} = dx \]

### Step 3: Integrate Both Sides
Now we will integrate both sides. The left side requires the integral of \( \frac{1}{1 + y^2} \), which is \( \tan^{-1}(y) \):
\[ \int \frac{dy}{1 + y^2} = \int dx \]
This gives us:
\[ \tan^{-1}(y) = x + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for the Constant of Integration
We use the initial condition \( y(0) = 0 \) to find \( C \):
\[ \tan^{-1}(0) = 0 + C \implies 0 = C \]
Thus, \( C = 0 \).

### Step 5: Write the Final Equation
Substituting \( C \) back into the equation, we have:
\[ \tan^{-1}(y) = x \]

### Step 6: Solve for \( y \)
Now we solve for \( y \):
\[ y = \tan(x) \]

### Conclusion
The solution to the differential equation \( y' = 1 + y^2 \) with the initial condition \( y(0) = 0 \) is:
\[ y = \tan(x) \]