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

Options

A. `y^(2)=exp(x+(x^(2))/(2))-1`
B. `y^(2)=1+Cexp(x+(x^(2))/(2))`
C. `y=tan(C+x+x^(2))`
D. `y=tan(x+(x^(2))/(2))`

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 can be rewritten as:
\[ y' = 1 + x + y^2(1 + x) \]
This allows us to factor out \( y^2 \):
\[ y' = 1 + x + y^2(1 + x) \]

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

### Step 3: 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 4: Apply the initial condition
Now we apply the initial condition \( y(0) = 0 \):
\[ \tan^{-1}(0) = 0 + \frac{0^2}{2} + C \implies 0 = C \]
So, \( C = 0 \).

### Step 5: Write the final solution
Substituting \( C \) back into the equation gives:
\[ \tan^{-1}(y) = x + \frac{x^2}{2} \]
To solve for \( y \), we take the tangent of both sides:
\[ y = \tan\left(x + \frac{x^2}{2}\right) \]

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

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