Prove a formula for additional solutions of a Riccati equation given one solution

Consider a differential equation

$y' + P(x) y + Q(x) y^2 = R(x)$

called a Riccati equation. Prove that if $u$ is a solution of this equation then there are additional solutions

$y = u + \frac{1}{v}$

where $v$ satisfies a first-order linear differential equation.

Proof. Let $u$ be a function satisfying

$u' + P(x) u + Q(x)u^2 = R(x).$

Further, let $y = u + \frac{1}{v}$ where $v$ satisfies a first-order linear differential equation. Then,

$y = u + \frac{1}{v} \quad \implies \quad y' = u' - \frac{v'}{v^2}.$

Therefore,

$\begin{align*} y' + P(x)y + Q(x) y^2 &= \left( u' - \frac{v'}{v^2} \right) + \left( u + \frac{1}{v} \right)P(x) + \left( u + \frac{1}{v} \right)^2 Q(x) \\[9pt] &= u' - \frav{v'}{v^2} + P(x) u + P(x) \frac{1}{v} + Q(x) u^2 + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2} \\[9pt] &= R(x) - \frac{v'}{v^2} + P(x)\frac{1}{v} + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2}. \end{align*}$

Thus, if

$-\frac{v'}{v^2} + P(x)v + 2Q(x) \frac{v}{u} + Q(x) \frac{1}{v^2} = 0,$

then $y = u+ \frac{1}{v}$ is a solution of our equation. But,

$\begin{align*} &&-\frac{v'}{v^2} + P(x) \frac{1}{v} + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2} &= 0 \\[9pt] \implies && -v' + P(x) v + 2Q(x) uv + Q(x) &= 0 \\[9pt] \implies && v' - v (P(x) + 2Q(x) u) &= Q(x). \end{align*}$

This is a first-order linear differential equation (since $u$ is a function of $x$ ). Hence, $y = u + \frac{1}{v}$ is a solution of

$y' + P(x) y + Q(x)y^2 = R(x)$

where $v$ is the solution of the first-order linear differential equation

$v' - v (P(x) + 2Q(x) u) = Q(x). \qquad \blacksquare$

Stumbling Robot

Prove a formula for additional solutions of a Riccati equation given one solution

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