📂Odinary Differential Equations

Riccati Differential Equation Solutions

Definition

The first-order nonlinear differential equation below is called the Riccati equation.

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

Explanation

If $y_{1}$ is known as a particular solution, the general solution is represented in the form of $y=y_{1}+u(x)$. Here, $u(x)$ is an arbitrary constant, and it can be obtained by solving the Bernoulli differential equation when $n=2$.

Solution

The Riccati equation looks too complicated at first glance to solve. Hence, we need to use a simple trick to change it into a form that is easier for us to solve.

• Step 1.

  Let’s say $y_{1}$ is a random particular solution. The way to obtain this particular solution is not fixed; it must be acquired through intuition, insight, or multiple attempts. Maybe by inserting $y$ into $2$? Maybe by inserting $3x$? And so on. And let’s assume the general solution is in the form of $y=y_{1}+u(x)$. $u(x)$ is an arbitrary constant.

• Step 2.

  When substituting $y=y_{1}+u$ into the given differential equation,

  $$\begin{align*} && y_{1}^\prime + u^\prime = P(y_{1}+u) + Q(y_{1}+u)^2 +R \\ \implies && y_{1}^\prime + u^\prime = P(y_{1}+u) + Q(y_{1}^2 +2y_{1}u +u^2) +R \end{align*}$$

  Organizing the right-hand side gives

  $$y_{1}^\prime + u^\prime = \left( Py_{1} + Qy_{1}^2 + R \right) +\left( P + 2Qy_{1} \right) u +Qu^2$$

• Step 3.

  Since $y_{1}$ is a solution to the given differential equation, it satisfies $y_{1}^\prime=Py_{1}+Qy_{1}^2+R$. Therefore, by canceling out the common terms on both sides, we get the following.

  $$u^\prime = \left( P + 2Qy_{1} \right) u +Qu^2$$

  This is the same form as when $n=2$ in Bernoulli’s differential equation. Therefore, $u(x)$ can be found using the solution method of Bernoulli’s differential equation. After that, the general solution $y=y_{1}+u(x)$ of the given differential equation is obtained.

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Example

Solve the differential equation $y^\prime = 2- 2xy+y^2$.

Since substituting $2x$ into $y$ works, we can assume $y_{1}=2x$. Then, the general solution is

$$y=y_{1}+u(x)=2x+u \\ y^\prime=2+u^\prime$$

Substituting into the given differential equation gives

$$\begin{align*} && 2+u^\prime &= 2 – 2x(2x+u)+(2x+u)^2 \\ \implies && 2+u^\prime &= 2-4x^2-2xu+4x^2+4xu+u^2 \\ \implies && u^\prime &= 2xu+u^2 \\ \implies && u^\prime –2xu &= u^2 \end{align*}$$

In other words, it is the form when $n=2$ in Bernoulli’s equation.

Bernoulli’s equation

$$u^\prime + (1-n)pu=q(1-n)$$

Dividing both sides by $u^2$ gives

$$u^{-2}u^\prime – 2xu^{-1}=1$$

By substituting $w \equiv u^{1-n}=u^{-1}$ and solving Bernoulli’s equation, we get $\dfrac{dw}{dx}=-u^{-2}\dfrac{du}{dx}$, so

$$\begin{align*} && -w^\prime –2xw &= 1 \\ \implies && w^\prime + 2xw &= -1 \\ \implies && w &= e^{-x^2} \left[ -\displaystyle \int e^{x^2}dx+C \right] \end{align*}$$

However, since we substituted $w =u^{-1}$ above,

$$u=\dfrac{e^{x^2}}{C- \displaystyle \int e^{x^2} dx}$$

Therefore, the final general solution is as follows.

$$y=y_{1}+u(x)=2x+\dfrac{e^{x^2}}{C- \displaystyle \int e^{x^2}dx}$$

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