What is the solution of `y'=1+x+y^(2)+xy^(2),y(0)=0` ?

Options

A. `y=tan^(2)((x^(2))/(2)+x)`
B. `y=tan^(2)(x^(2)+x)`
C. `y=tan(x^(2)+x)`
D. `y=tan((x^(2))/(2)+x)`

Answer

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

### Step 1: Rewrite the equation
The given equation is:
\[ y' = 1 + x + y^2 + xy^2 \]
We can express \( y' \) as \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 1 + x + y^2 + xy^2 \]

### Step 2: Rearrange the equation
We can factor out \( y^2 \):
\[ \frac{dy}{dx} = 1 + x + y^2(1 + x) \]

### Step 3: Separate variables
Now, we can separate the variables \( y \) and \( x \):
\[ \frac{dy}{1 + y^2} = (1 + x)dx \]

### Step 4: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dy}{1 + y^2} = \int (1 + x)dx \]
The left side integrates to:
\[ \tan^{-1}(y) \]
The right side integrates to:
\[ x + \frac{x^2}{2} + C \]
Thus, we have:
\[ \tan^{-1}(y) = x + \frac{x^2}{2} + C \]

### Step 5: Solve for \( y \)
To find \( y \), we take the tangent of both sides:
\[ y = \tan\left(x + \frac{x^2}{2} + C\right) \]

### Step 6: Apply the initial condition
We apply the initial condition \( y(0) = 0 \):
\[ 0 = \tan\left(0 + \frac{0^2}{2} + C\right) \implies C = 0 \]
So the equation simplifies to:
\[ y = \tan\left(x + \frac{x^2}{2}\right) \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = \tan\left(x + \frac{x^2}{2}\right) \]

---