Determine the particular solution that satisfies the differential equation and initial condition below. $y^{\prime}=x^{2}-10 x-14$ where $y=3$ when $x=0$ \[ y= \]

Step 1 ：The given differential equation is a first order linear differential equation. The general solution of this differential equation can be obtained by integrating the right hand side of the equation.

Step 2 ：The general solution is \(y = C + \frac{x^{3}}{3} - 5x^{2} - 14x\), where C is the constant of integration.

Step 3 ：The particular solution can then be obtained by substituting the initial condition into the general solution and solving for the constant of integration.

Step 4 ：The initial condition is y(0) = 3, which means when x = 0, y = 3. This should be used to solve for the constant of integration C in the general solution.

Step 5 ：Substituting x = 0 into the general solution gives C = 3.

Step 6 ：Substituting C = 3 into the general solution gives the particular solution \(y = \frac{x^{3}}{3} - 5x^{2} - 14x + 3\).

Step 7 ：Final Answer: The particular solution that satisfies the differential equation and initial condition is \(\boxed{\frac{x^{3}}{3} - 5x^{2} - 14x + 3}\)