# Riccati Equations Before we give the formal definition of Riccati equations, a little introduction may be helpful. Indeed, consider the first order differential equation If we approximate f(x,y), while x is kept constant, we will get If we stop at y, we will get a linear equation. Riccati looked at the approximation to the second degree: he considered equations of the type These equations bear his name, Riccati equations. They are nonlinear and do not fall under the category of any of the classical equations. In order to solve a Riccati equation, one will need a particular solution. Without knowing at least one solution, there is absolutely no chance to find any solutions to such an equation. Indeed, let y1 be a particular solution of Consider the new function z defined by Then easy calculations give which is a linear equation satisfied by the new function z. Once it is solved, we go back to y via the relation Keep in mind that it may be harder to remember the above equation satisfied by z. Instead, try to do the calculations whenever you can.

Example. Solve the equation knowing that y1 = 2 is a particular solution.

Answer. We recognize a Riccati equation. First of all we need to make sure that y1 is indeed a solution. Otherwise, our calculations will be fruitless. In this particular case, it is quite easy to check that y1 = 2 is a solution. Set Then we have which implies Hence, from the equation satisfied by y, we get Easy algebraic manipulations give Hence

z' = -3z -1.

This is a linear equation. The general solution is given by Therefore, we have Note: If one remembers the equation satisfied by z, then the solutions may be found a bit faster. Indeed in this example, we have P(x) = -2, Q(x) = -1, and R(x) = 1. Hence the linear equation satisfied by the new function z, is Example. Check that is a solution to Then solve the IVP We will let the reader check that is indeed a particular solution of the given differential equations. We also recognize that the equation is of Riccati type. Set which gives Hence Substituting into the equation gives Easy algebraic manipulations give Hence This is the linear equation satisfied by z. The integrating factor is The general solution is Now it is time to go back to the original function y. We have The initial condition y(0) = -1 implies 1/C = -1, or C = -1. Therefore the solution to the IVP is  [Differential Equations] [Algebra] [Trigonometry ]
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Last Update 9-15-98