How do you solve $\dfrac{{dy}}{{dx}} = x{y^2}$ ?

Hint: We have been given to solve a differential equation. A differential equation is an equation which represents the derivative of an independent variable with respect to a dependent variable. A differential equation containing only the first order of derivatives is known as a first order differential equation. To solve the given equation we can use the variable separable method which works by separating both the variables to different sides of the equation. The general solution of a first order differential equation is a relation between the variables obtained after eliminating or simplifying the derivatives.

Complete step by step solution:
We have to find the solution of the equation $\dfrac{{dy}}{{dx}} = x{y^2}$.
This is a first order differential equation whose solution will be obtained after eliminating the derivative term. For this we will use the variable separable method.
In the variable separable method we try to separate the terms of a particular variable on each side of the equation.
We can rearrange the given equation to write it as,
$
  \dfrac{{dy}}{{dx}} = x{y^2} \\
   \Rightarrow \dfrac{{dy}}{{{y^2}}} = xdx \\
 $
We can see that all terms with variable $y$ is on the LHS and with variable $x$ is on the RHS.
Now we integrate both sides and simplify.
$
  \int {\dfrac{{dy}}{{{y^2}}}} = \int {xdx} \\
   \Rightarrow \dfrac{{{y^{ - 2 + 1}}}}{{ - 2 + 1}} = \dfrac{{{x^{1 + 1}}}}{{1 + 1}} + c \\
   \Rightarrow - \dfrac{1}{y} = \dfrac{{{x^2}}}{2} + c \\
   \Rightarrow \dfrac{{{x^2}}}{2} + \dfrac{1}{y} = C \\
 $
We can observe that we have eliminated the derivative terms and got the resulting expression as $\dfrac{{{x^2}}}{2} + \dfrac{1}{y} = C$.

Hence, the solution to $\dfrac{{dy}}{{dx}} = x{y^2}$ is $\dfrac{{{x^2}}}{2} + \dfrac{1}{y} = C$, where $C$ is any arbitrary constant.

Note: We have used a variable separable method to solve the given first order differential equation. This method is applicable only when we are able to separate the terms of a particular variable on each side of the equation. In the solution we are left with no terms of derivatives. We add an arbitrary constant in the solution as we are solving an indefinite integral in the process.