Differential Equations

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Riccati Equation

General Riccati Equation

The Riccati equation is one of the most interesting nonlinear differential equations of first order. It's written in the form:

\[y' = a\left( x \right)y + b\left( x \right){y^2} + c\left( x \right),\]

where \(a\left( x \right),\) \(b\left( x \right),\) \(c\left( x \right)\) are continuous functions of \(x.\)

The Riccati equation is used in different areas of mathematics (for example, in algebraic geometry and the theory of conformal mapping), and physics. It also appears in many applied problems.

The differential equation given above is called the general Riccati equation. It can be solved with help of the following theorem:

Theorem.

If a particular solution \({y_1}\) of a Riccati equation is known, the general solution of the equation is given by

\[y = {y_1} + u.\]

Indeed, substituting the solution \(y = {y_1} + u\) into Riccati equation, we have

\[\left( {{y_1} + u} \right)^\prime = a\left( x \right)\left( {{y_1} + u} \right) + b\left( x \right){\left( {{y_1} + u} \right)^2} + c\left( x \right),\]
\[\underline {{y_1}^\prime } + u' = \underline {a\left( x \right){y_1}} + a\left( x \right)u + \underline {b\left( x \right)y_1^2} + 2b\left( x \right){y_1}u + b\left( x \right){u^2} + \underline {c\left( x \right)} .\]

The underlined terms in the left and in the right side can be canceled because \({y_1}\) is a particular solution satisfying the equation. As a result we obtain the differential equation for the function \(u\left( x \right):\)

\[u' = b\left( x \right){u^2} + \left[ {2b\left( x \right){y_1} + a\left( x \right)} \right]u,\]

which is a Bernoulli equation.

Substitution of \(z = \frac{1}{u}\) converts the given Bernoulli equation into a linear differential equation that allows integration.

Besides the general Riccati equation, there is an infinite number of particular cases of Riccati equation at certain coefficients of \(a\left( x \right),\) \(b\left( x \right),\) and \(c\left( x \right).\) Many of these particular cases have integrable solutions.

Returning to the general Riccati equation, we see that we can construct the general solution if a particular solution is known. Unfortunately, there is no strict algorithm to find the particular solution, which depends on the types of the functions \(a\left( x \right),\) \(b\left( x \right),\) and \(c\left( x \right).\)

Below we consider some well known particular cases of the Riccati equation.

Special Case \(1:\) Coefficients \(a, b, c\) are constants

If the coefficients in the Riccati equation are constants, this equation can be reduced to a separable differential equation. The solution is described by the integral of a rational function with a quadratic function in the denominator:

\[y' = ay + b{y^2} + c,\;\; \Rightarrow \frac{{dy}}{{dx}} = ay + b{y^2} + c,\;\; \Rightarrow \int {\frac{{dy}}{{ay + b{y^2} + c}}} = \int {dx} .\]

This integral can be easily calculated at any values of \(a,\) \(b\) and \(c\) (For more information see Integration of Rational Functions).

Special Case \(2:\) Equation of type \(y' = b{y^2} + c{x^n}\)

Consider a Riccati equation of type \(y' = b{y^2} + c{x^n},\) where the function \(a\left( x \right)\) at the linear term is zero, the coefficient \(b\) at \({y^2}\) is a constant, and \(c\left( x \right)\) is a power function:

\[a\left( x \right) \equiv 0,\;\; b\left( x \right) = b,\;\; c\left( x \right) = c{x^n}.\]

This case of Riccati equation has nice solutions!

First of all, if \(n = 0,\) we get the Case \(1\) where the variables are separated and the differential equation can be integrated.

If \(n = -2,\) the Riccati equation is converted into a homogeneous equation with help of the substitution \(y = \frac{1}{z}\) and then also can be integrated.

This differential equation can be also solved at

\[n = \frac{{4k}}{{1 - 2k}},\;\; \text{where}\;\; k = \pm 1, \pm 2, \pm 3, \ldots \]

Here the general solution is expressed through cylinder functions.

At all other values of the power \(n,\) the solution of the Riccati equation can be expressed through integrals of elementary functions. This fact was discovered by the French mathematician Joseph Liouville \(\left( {1809 - 1882} \right)\) in \(1841.\)

Solved Problems

Click or tap a problem to see the solution.

Example 1

Solve the differential equation \[y' = y + {y^2} + 1.\]

Example 2

Solve the Riccati equation \[y' + {y^2} = \frac{2}{{{x^2}}}.\]

Example 1.

Solve the differential equation \[y' = y + {y^2} + 1.\]

Solution.

The given equation is a simple Riccati equation with constant coefficients. Here the variables \(x, y\) can be easily separated, so the general solution of the equation is given by

\[\frac{{dy}}{{dx}} = y + {y^2} + 1,\;\; \Rightarrow \frac{{dy}}{{y + {y^2} + 1}} = dx,\;\; \Rightarrow \int {\frac{{dy}}{{y + {y^2} + 1}}} = \int {dx} ,\;\; \Rightarrow \int {\frac{{dy}}{{{y^2} + y + \frac{1}{4} + \frac{3}{4}}}} = \int {dx} ,\;\; \Rightarrow \int {\frac{{dy}}{{{{\left( {y + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = \int {dx} ,\;\; \Rightarrow \frac{1}{{\frac{{\sqrt 3 }}{2}}}\arctan \frac{{y + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = x + C,\;\; \Rightarrow \frac{2}{{\sqrt 3 }}\arctan \frac{{2y + 1}}{{\sqrt 3 }} = x + C.\]

Example 2.

Solve the Riccati equation \[y' + {y^2} = \frac{2}{{{x^2}}}.\]

Solution.

We will seek for a particular solution in the form:

\[y = \frac{c}{x},\;\; \Rightarrow y' = - \frac{c}{{{x^2}}}.\]

Substituting this into the equation, we obtain:

\[- \frac{c}{{{x^2}}} + {\left( {\frac{c}{x}} \right)^2} = \frac{2}{{{x^2}}}\;\; \text{or}\;\; - \frac{c}{{{x^2}}} + \frac{{{c^2}}}{{{x^2}}} = \frac{2}{{{x^2}}}.\]

We get a quadratic equation for \(c:\)

\[{c^2} - c - 2 = 0,\;\; \Rightarrow D = 1 - 4 \cdot \left( { - 2} \right) = 9,\;\; \Rightarrow {c_{1,2}} = \frac{{1 \pm 3}}{2} = - 1,2.\]

We can take any value of\(c.\) For example, let \(c = 2.\) Now, when the particular solution is known, we make the replacement:

\[y = z + \frac{2}{x},\;\; \Rightarrow y' = z' - \frac{2}{{{x^2}}}.\]

Now substitute this into the original Riccati equation:

\[z' - \frac{2}{{{x^2}}} + {\left( {z + \frac{2}{x}} \right)^2} = \frac{2}{{{x^2}}},\;\; \Rightarrow z' - \cancel{\frac{2}{{{x^2}}}} + {z^2} + \frac{4}{x}z + \cancel{\frac{4}{{{x^2}}}} = \cancel{\frac{2}{{{x^2}}}},\;\; \Rightarrow z' + \frac{4}{x}z = - {z^2}.\]

As it can be seen, we have a Bernoulli equation with the parameter \(m = 2.\) Make one more substitution:

\[v = {z^{1 - m}} = \frac{1}{z},\;\; \Rightarrow v' = - \frac{{z'}}{{{z^2}}}.\]

Divide the Bernoulli equation by \({z^2}\) (assuming that \(z \ne 0\)) and rewrite it in terms of \(v:\)

\[\frac{{z'}}{{{z^2}}} + \frac{{4z}}{{x{z^2}}} = - 1,\;\; \Rightarrow - \frac{{z'}}{{{z^2}}} - \frac{4}{{xz}} = 1,\;\; \Rightarrow v' - \frac{4}{x}v = 1.\]

The last equation is linear and can be easily solved using the integrating factor:

\[u = {e^{\int {\left( { - \frac{4}{x}} \right)dx} }} = {e^{ - 4\int {\frac{{dx}}{x}} }} = {e^{ - 4\ln \left| x \right|}} = {e^{\ln \frac{1}{{{{\left| x \right|}^4}}}}} = \frac{1}{{{{\left| x \right|}^4}}} = \frac{1}{{{x^4}}}.\]

The general solution of the linear equation is given by

\[v = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {\frac{1}{{{x^4}}} \cdot 1dx} + C}}{{\frac{1}{{{x^4}}}}} = \frac{{\int {{x^{ - 4}}dx} + C}}{{\frac{1}{{{x^4}}}}} = \left( { - \frac{1}{3}{x^{ - 3}} + C} \right){x^4} = - \frac{x}{3} + C{x^4}.\]

From now on we will subsequently return back to the previous variables. Since \(z = \frac{1}{v},\) the general solution for \(z\) is written as follows:

\[\frac{1}{z} = - \frac{x}{3} + C{x^4},\;\; \Rightarrow z = \frac{1}{{ - \frac{x}{3} + C{x^4}}} = - \frac{3}{{x + 3C{x^4}}} = - \frac{3}{{x\left( {1 + 3C{x^3}} \right)}}.\]

Hence,

\[y = z + \frac{2}{x} = - \frac{3}{{x\left( {1 + 3C{x^3}} \right)}} + \frac{2}{x} = \frac{{ - 3 + 2\left( {1 + 3C{x^3}} \right)}}{{x\left( {1 + 3C{x^3}} \right)}} = \frac{{ - 3 + 2 + 6C{x^3}}}{{x\left( {1 + 3C{x^3}} \right)}} = \frac{{6C{x^3} - 1}}{{x\left( {1 + 3C{x^3}} \right)}}.\]

We can rename the constant: \(3C = {C_1}\) and write the answer in the form:

\[y = \frac{{2{C_1}{x^3} - 1}}{{x\left( {1 + {C_1}{x^3}} \right)}},\]

where \({C_1}\) is an arbitrary real number.

See more problems on Page 2.

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