# 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

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