Riccati's Equation

Riccati's equation is the simplest genuinely nonlinear first-order equation, and in general it cannot be solved in closed form. But it has a striking property: if you can find one particular solution by any means, however lucky the guess, the substitution $y = y_{1} + 1 / u$ reduces the equation to a linear one and delivers the complete general solution.

Quick Reference

Item	Statement
Riccati's equation	$\frac{d y}{d x} + P (x) y + Q (x) y^{2} = f (x)$
Nonlinear unless	$Q (x) \equiv 0$ , in which case it is linear
Requirement	One particular solution $y_{1} (x)$ must be known
Bernoulli route	$y = y_{1} + z$ gives z' + \left[P + 2Qy_1\right]z + Qz^{2} = 0
Direct route	$y = y_{1} + \frac{1}{u}$ gives the linear equation u' - \left[P(x) + 2Q(x)y_1\right]u = Q(x)
Final answer	$y = y_{1} (x) + \frac{1}{u (x)}$
Finding $y_{1}$	By inspection: try constants, $a x + b$ , or $a / x$ , guided by the shape of $P$ , $Q$ , and $f$
Related equation	Clairaut's equation y = xy' + g(y'), a special case of the Lagrange equation

The Equation

An equation of the form

\frac{d y}{d x} + P (x) y + Q (x) y^{2} = f (x)

is called Riccati's equation.

This is a nonlinear equation unless $Q (x) \equiv 0$ , in which case it reduces to the linear equation of Section 2.6. The single term $Q (x) y^{2}$ is what removes it from the reach of every linear method.

If a single particular solution $y_{1} (x)$ is known, then the equation can be solved easily. Everything below depends on having such a solution in hand.

Method 1: Reduce to Bernoulli's Equation

The equation can be transformed into Bernoulli's equation by a change of variable. Assume

y (x) = y_{1} (x) + z (x)

where $z$ is the new unknown, measuring the deviation of $y$ from the known solution. Substituting,

y_1' + z' + P(x)\left(y_1 + z\right) + Q(x)\left(y_1 + z\right)^{2} = f(x) .

Expanding the square, ${(y_{1} + z)}^{2} = y_{1}^{2} + 2 y_{1} z + z^{2}$ , and grouping the terms that involve $y_{1}$ alone,

\underbrace{\left[y_1' + P(x)y_1 + Q(x)y_1^{2}\right]}_{=\ f(x)} + z' + \left[P(x) + 2Q(x)y_1\right]z + Q(x)z^{2} = f(x) .

Because $y_{1}$ is a solution, the bracketed group equals $f (x)$ and cancels against the right-hand side. What remains is

\frac{d z}{d x} + [P (x) + 2 Q (x) y_{1}] z + Q (x) z^{2} = 0,

a Bernoulli equation with $α = 2$ .

To solve a Bernoulli equation with $α = 2$ we substitute $u = z^{- 1}$ . So from the beginning we could have used the single substitution

y (x) = y_{1} (x) + \frac{1}{u (x)}

to transform the equation directly into a linear equation.

Method 2: Go Directly to a Linear Equation

Take $y = y_{1} + \frac{1}{u}$ , so that by the chain rule

y' = y_1' - \frac{1}{u^{2}}u' .

Substituting into Riccati's equation,

y_1' - \frac{u'}{u^{2}} + P(x)\left[y_1 + \frac{1}{u}\right] + Q(x)\left[y_1^{2} + \frac{2y_1}{u} + \frac{1}{u^{2}}\right] = f(x) .

Rearranging so that the terms in $y_{1}$ alone are grouped together,

\underbrace{\left[y_1' + Py_1 + Qy_1^{2}\right]}_{=\ f(x)} + \left(-\frac{u'}{u^{2}} + \frac{P + 2Qy_1}{u} + \frac{Q}{u^{2}}\right) = f(x)

The underbraced group cancels against $f (x)$ , leaving

-\frac{u'}{u^{2}} + \frac{P + 2Qy_1}{u} + \frac{Q}{u^{2}} = 0 .

Multiplying by $u^{2}$ ,

-u' + \left[P(x) + 2Q(x)y_1\right]u + Q(x) = 0

and multiplying by $- 1$ to put the derivative first,

u' - \left[P(x) + 2Q(x)y_1(x)\right]u = Q(x) .

This is a linear first-order equation in $u$ , solved by the integrating factor

I (x) = \exp (- \int [P + 2 Q y_{1}] d x) .

Once $u$ is found, the general solution of Riccati's equation is $y = y_{1} + \frac{1}{u}$ .

Note where the arbitrary constant enters. The linear equation for $u$ contributes one constant, and $y = y_{1} + 1 / u$ carries it through to the answer. That is the correct count for a first-order equation, and it confirms that this method finds the general solution, not just another particular one.

Finding a Particular Solution

There is no general procedure, which is the honest limitation of the method. In practice you guess a form suggested by the coefficients and check it:

If $P$ , $Q$ , and $f$ are all constants, try a constant $y_{1} = k$ and solve the resulting quadratic.
If the coefficients are polynomials of low degree, try $y_{1} = a x + b$ and match coefficients.
If the coefficients involve $1 / x$ or $1 / x^{2}$ , try $y_{1} = a / x$ .
If $f (x)$ resembles $Q (x)$ times something squared, that something is often the particular solution.

Worked Examples

Solve y' = 1 + x^{2} - 2xy + y^{2}, given that $y_{1} = x$ is a particular solution.

Solution

First put the equation in the standard Riccati form by moving everything but y' to the left:

y' + 2xy - y^{2} = 1 + x^{2} ,

so $P (x) = 2 x$ , $Q (x) = - 1$ , and $f (x) = 1 + x^{2}$ .

Verify that $y_{1} = x$ works: y_1' = 1, and

1 + 2 x \cdot x - x^{2} = 1 + 2 x^{2} - x^{2} = 1 + x^{2} = f (x) . ✓

Now apply the linear equation for $u$ :

u' - \left[P + 2Qy_1\right]u = Q \quad \Longrightarrow \quad u' - \left[2x + 2(-1)(x)\right]u = -1 .

The bracket collapses to $2 x - 2 x = 0$ , so

u' = -1 \quad \Longrightarrow \quad u = c - x .

Therefore the general solution is

y = x + \frac{1}{c - x} .

Check. Differentiating, y' = 1 + \dfrac{1}{(c-x)^{2}}. On the other side, note that the original equation can be written y' = 1 + (x - y)^{2}, and $x - y = - \frac{1}{c - x}$ , so $(x - y)^{2} = \frac{1}{(c - x)^{2}}$ . Both sides agree.

Solve y' = y^{2} - \dfrac{2}{x^{2}} for $x > 0$ .

Solution

Step 1: find a particular solution. The coefficient $2 / x^{2}$ suggests trying $y_{1} = \frac{a}{x}$ . Then y_1' = -\dfrac{a}{x^{2}} and the equation demands

- \frac{a}{x^{2}} = \frac{a^{2}}{x^{2}} - \frac{2}{x^{2}} ⟹ - a = a^{2} - 2 ⟹ a^{2} + a - 2 = 0 .

The roots are $a = 1$ and $a = - 2$ . Take $y_{1} = \frac{1}{x}$ .

Step 2: identify $P$ and $Q$ . Writing the equation in standard form,

y' + 0\cdot y - y^{2} = -\frac{2}{x^{2}} ,

so $P = 0$ and $Q = - 1$ .

Step 3: solve the linear equation.

u' - \left[0 + 2(-1)\frac{1}{x}\right]u = -1 \quad \Longrightarrow \quad u' + \frac{2}{x}u = -1 .

The integrating factor is $I = e^{\int (2 / x) d x} = x^{2}$ , so

\frac{d}{d x} (x^{2} u) = - x^{2} ⟹ x^{2} u = - \frac{x^{3}}{3} + c ⟹ u = \frac{c}{x^{2}} - \frac{x}{3} .

Step 4: assemble.

y = \frac{1}{x} + \frac{1}{\frac{c}{x^{2}} - \frac{x}{3}} = \frac{1}{x} + \frac{3 x^{2}}{3 c - x^{3}} .

Check with $c = 0$ . Then $y = \frac{1}{x} + \frac{3 x^{2}}{- x^{3}} = \frac{1}{x} - \frac{3}{x} = - \frac{2}{x}$ , which is the other particular solution found in Step 1. Verifying: y' = \dfrac{2}{x^{2}} and $y^{2} - \frac{2}{x^{2}} = \frac{4}{x^{2}} - \frac{2}{x^{2}} = \frac{2}{x^{2}}$ . Both agree.

Clairaut's Equation

A different nonlinear first-order equation is worth knowing, because it is one of the few whose singular solutions can be found systematically.

Clairaut's equation has the form

y = x y' + g(y')

for a given function $g$ . It is a particular case of the Lagrange equation y = x\,h(y') + g(y').

Write p = y', so the equation reads $y = x p + g (p)$ . Differentiating both sides with respect to $x$ ,

p = p + x\frac{dp}{dx} + g'(p)\frac{dp}{dx} \quad \Longrightarrow \quad \left[x + g'(p)\right]\frac{dp}{dx} = 0 .

A product is zero when one of its factors is, which splits the problem into two cases.

Case 1: $\frac{d p}{d x} = 0$ . Then $p = c$ is constant, and substituting back gives the general solution

y = c x + g (c),

a one-parameter family of straight lines.

Case 2: x + g'(p) = 0. Solving this for $p$ and substituting into $y = x p + g (p)$ gives the singular solution, which is the envelope of the family of lines. It carries no arbitrary constant and is not obtainable from the general solution for any value of $c$ .

Solve Clairaut's equation y = xy' - \left(y'\right)^{2}.

Solution

Here $g (p) = - p^{2}$ , so g'(p) = -2p.

General solution. Setting $p = c$ ,

y = c x - c^{2},

a family of straight lines.

Singular solution. From x + g'(p) = 0 we get $x - 2 p = 0$ , so $p = \frac{x}{2}$ . Substituting into $y = x p - p^{2}$ ,

y = x \cdot \frac{x}{2} - \frac{x^{2}}{4} = \frac{x^{2}}{2} - \frac{x^{2}}{4} = \frac{x^{2}}{4} .

So $y = \frac{x^{2}}{4}$ is a singular solution. It is the parabola tangent to every line $y = c x - c^{2}$ , touching the line of slope $c$ at the point $x = 2 c$ . No choice of $c$ produces this parabola, which is exactly what makes the solution singular.

Exercises

Solve y' = y^{2} - 2xy + x^{2} + 1, given that $y_{1} = x$ is a particular solution.

Solution

This is the same equation as the first worked example, since $y^{2} - 2 x y + x^{2} + 1 = 1 + x^{2} - 2 x y + y^{2}$ . The general solution is

y = x + \frac{1}{c - x} .

Solve y' + y^{2} = \dfrac{2}{x^{2}} for $x > 0$ , after finding a particular solution of the form $a / x$ .

Solution

Try $y_{1} = a / x$ : then $- \frac{a}{x^{2}} + \frac{a^{2}}{x^{2}} = \frac{2}{x^{2}}$ , so $a^{2} - a - 2 = 0$ , giving $a = 2$ or $a = - 1$ . Take $y_{1} = \frac{2}{x}$ .

In standard form $P = 0$ and $Q = 1$ , so

u' - \left[0 + 2\cdot 1\cdot\frac{2}{x}\right]u = 1 \quad \Longrightarrow \quad u' - \frac{4}{x}u = 1 .

The integrating factor is $x^{- 4}$ , giving \left(x^{-4}u\right)' = x^{-4}, so

x^{- 4} u = - \frac{1}{3} x^{- 3} + c ⟹ u = - \frac{x}{3} + c x^{4} .

Therefore

y = \frac{2}{x} + \frac{1}{c x^{4} - \frac{x}{3}} = \frac{2}{x} + \frac{3}{3 c x^{4} - x} .

Taking $c = 0$ recovers $y = \frac{2}{x} - \frac{3}{x} = - \frac{1}{x}$ , the second particular solution.

Find the general and singular solutions of y = xy' + \dfrac{1}{y'}.

Solution

Here $g (p) = \frac{1}{p}$ , so g'(p) = -\dfrac{1}{p^{2}}.

General solution: $y = c x + \frac{1}{c}$ for $c \neq 0$ .

Singular solution: $x - \frac{1}{p^{2}} = 0$ gives $p = \pm \frac{1}{\sqrt{x}}$ for $x > 0$ . Substituting the positive root,

y = x \cdot \frac{1}{\sqrt{x}} + \sqrt{x} = \sqrt{x} + \sqrt{x} = 2 \sqrt{x} .

So $y = 2 \sqrt{x}$ , equivalently $y^{2} = 4 x$ , a parabola. The negative root gives $y = - 2 \sqrt{x}$ , the other branch of the same parabola.

Frequently Asked Questions

What is Riccati's equation?

It is the first-order equation y' + P(x)y + Q(x)y^{2} = f(x), that is, a linear equation with one extra quadratic term. That single term makes it nonlinear, and it is the simplest nonlinear equation for which a systematic theory exists.

Why do I need a particular solution first?

Because Riccati's equation cannot be solved in closed form in general. There is no formula analogous to the linear case. What the theory offers instead is a conditional result: given one solution, the substitution $y = y_{1} + 1 / u$ produces a linear equation, and the linear equation always solves. Without $y_{1}$ , the substitution has nothing to build on.

How do I find that first solution?

By inspection, guided by the shape of the coefficients. Constant coefficients suggest a constant $y_{1}$ ; polynomial coefficients suggest $y_{1} = a x + b$ ; terms in $1 / x^{2}$ suggest $y_{1} = a / x$ . Substitute the trial form, match coefficients, and solve for the parameters. Textbook problems almost always supply $y_{1}$ or make it findable this way.

Does it matter which particular solution I start from?

No. Different choices of $y_{1}$ lead to different intermediate equations for $u$ , but the same family of solutions at the end, with the constant relabeled. In the second worked example, starting from $y_{1} = - 2 / x$ instead of $1 / x$ produces the same family, as the check with $c = 0$ illustrates.

Why does the substitution

y = y_{1} + 1 / u

eliminate the nonlinearity?

The quadratic term expands as $Q (y_{1}^{2} + 2 y_{1} / u + 1 / u^{2})$ . The $y_{1}^{2}$ piece is absorbed by the fact that $y_{1}$ solves the equation, and the remaining $1 / u^{2}$ piece is cancelled after multiplying through by $u^{2}$ , since the derivative term contributes -u'/u^{2} with the same denominator. Nothing quadratic in $u$ survives.

How is Clairaut's equation related to Riccati's?

Only loosely. Both are nonlinear first-order equations with special structure. Clairaut's equation is included here because it is the standard example of an equation whose singular solution can be computed systematically, as the envelope of the general solution's family of lines, which connects back to the discussion of singular solutions in Section 1.2.

Quick Reference

The Equation

Method 1: Reduce to Bernoulli's Equation

Method 2: Go Directly to a Linear Equation

Finding a Particular Solution

Worked Examples

Clairaut's Equation

Exercises

Frequently Asked Questions

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