Solution of \((\mathrm{dy} / \mathrm{dx})=1+\mathrm{x}+\mathrm{y}^{2}+\mathrm{xy}^{2}, \mathrm{y}(0)=0\) is: (A) \(y=\tan \left(c+x+x^{2}\right)\) (B) \(\mathrm{y}=\tan \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\) (C) \(\mathrm{y}^{2}=\exp \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\) (D) \(\mathrm{y}^{2}=1+\mathrm{c} \cdot \exp \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\)

First order differential equations are a fundamental category of differential equations characterized by derivatives of the first degree. These equations involve functions and their first derivatives. The general form of a first order differential equation can be written as \(\frac{dy}{dx}=f(x,y)\), where \(f(x,y)\) is a function that can contain both the independent variable \(x\) and the dependent variable \(y\).

When solving a first order differential equation, one aims to find the function \(y(x)\) that satisfies the equation for given initial conditions. An essential technique for solving these equations is separation of variables, where the equation is manipulated to isolate \(dy\) and \(dx\) on different sides, allowing integration to find the solution. However, not all first order equations can be solved with this method, and other techniques such as integrating factors, exact equations, or substitution may be required.

For the given exercise, we have the first order differential equation \(\frac{dy}{dx}=1+x+y^2+xy^2\), which does not readily allow for separation of variables. The equation includes terms involving both \(x\) and \(y\), as well as nonlinear terms like \(y^2\) and \(xy^2\), which suggest that a direct integration method may not be applicable.

Initial value problems (IVPs) are specific types of differential equations where the solution is required to pass through a given point, typically specified as a condition on the dependent variable when the independent variable has a certain value. For instance, \(y(x_0)=y_0\) is an initial condition stating that at the point \(x_0\), the solution \(y\) must equal \(y_0\).

IVPs are crucial because they allow for a unique solution to a first order differential equation. This is a consequence of existence and uniqueness theorems which state, under certain conditions, that there is precisely one function \(y(x)\) that satisfies both the differential equation and the initial condition. Determining the function that meets these criteria can sometimes be challenging and may involve a variety of mathematical techniques.

In the exercise provided, the initial value problem is \(y(0)=0\). Here, the function \(y\), which is also the solution to our differential equation, must satisfy the condition that when \(x=0\), \(y\) must equal \(0\). This information is critical in eliminating possible solutions and in confirming the validity of the option that correctly represents the behavior of the solution as it passes through the initial point.

Integration of functions is a core concept in calculus, often used to find solutions to differential equations. It involves finding the antiderivative or integral of a function, which, in simple terms, is the opposite process of differentiation. Integrals can be either indefinite (without bounds) or definite (with bounds), the latter of which results in a numerical value representing the area under the curve of the function between two points.

In the context of differential equations, integration is used to calculate the function's original form from its derivative. This is especially pertinent for first order differential equations where, once separated, the terms involving \(dy\) and \(dx\) are integrated on their respective sides to find the function \(y\). However, for complex equations, such as non-separable ones, direct integration is not feasible, and alternative methods need to be employed.

The exercise under consideration does not yield to simple integration strategies because the relationship involving \(x\) and \(y\) is not easily separable. Rather, solving such an equation often requires more advanced techniques or clever manipulations to express the function in a form that allows for integration, or to reason by differentiation as demonstrated in the step-by-step solution to verify a suggested function as the correct solution.