One Parameter family of soultions for this Riccati's Equation

Question

Find a one parameter family of solutions of the Riccati's equation $$\frac{dy}{dx}=-\frac4{x^2}-\frac{y}{x}+y^2$$

So I am trying to follow a sort of template I was given:

Let $y=y_1+u$, where $y_1$ is a particular solution of the DE. Then, $$\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{du}{dx}$$

Plugging into DE and we get, $$\frac{dy_1}{dx}+\frac{du}{dx}=-\frac4{x^2}-\frac{y_1}{x}-\frac{u}{x}+y_1^2+2uy_1+u^2$$

Since $y_1$ is a particular solution, $$\frac{dy_1}{dx}=-\frac4{x^2}-\frac{y_1}{x}+y_1^2$$

Plugging in and canceling we get, $$\frac{du}{dx}=-\frac{u}{x}+2uy_1+u^2 \implies$$ $$\frac{du}{du}-u(\frac1x-2y_1)=u^2$$ Which is now a Bernoulli's Equation.

Question:

How can I complete this problem and find my 1-parameter family of solutions? Maybe I am thinking of this wrong, but I cannot do a Bernoulli's equation if $y_1$ is still there right? I believe I need to solve for $y_1$ but I am not sure how to.

Thank You for the help.

Answer 1

hint: With $$y(x)=\frac{v(x)}{x}$$ we get $$\frac{dv(x)}{dx}=\frac{v(x)^2-4}{x}$$ so we obtain $$\frac{\frac{dv(x)}{dx}}{v(x)^2-4}=\frac{1}{x}$$

Answer 2

The method that you use to solve a Riccati's equation is the simplest, insofar one know a particular solution.

In academic exercises the most likely one can guess a particular solution by inspection and check that the guess is fine. Then, the solving is easy. This is the case in the present exercise with the particular solution $y_1=\frac{2}{x}$ .

When one cannot find a particular solution, the method is as follows.

The Riccati's ODE is a non-linear ODE. It is well-known that a non-linear ODE is generally more difficult to solve that a linear ODE, even if the order of linear ODE is higher that the order of the non-linear ODE. That is why there is an advantage to transform the first order non-linear ODE into a second order linear ODE.

The method to transform a Riccati's ODE into a second order linear ODE is very simple. The general form of the Riccati's ODE is : $$y'(x)=P(x)+Q(x)y(x)+R(x)(y(x))^2$$ The change of function : $$y(x)=-\frac{f'(x)}{R(x)f(x)}$$ leads to a linear second order ODE.

Let try this method with the equation : $$\frac{dy}{dx}=-\frac4{x^2}-\frac{y}{x}+y^2$$ We see that $R(x)=1$, so the change of function is : $$y(x)=-\frac{f'(x)}{f(x)}$$ $$y'=-\frac{f''}{f}+\frac{f'^2}{f^2}=-\frac{4}{x^2}+\frac{f'}{xf}+\frac{f'^2}{f^2}$$ $$x^2f''+xf'-4f=0$$ It is easy to solve the linear second order ODE : $$f=c_1x^2+c_2\frac{1}{x^2}$$ $$f'=2c_1x-2c_2\frac{1}{x^3}$$ $$y=-\frac{2c_1x-2c_2\frac{1}{x^3}}{c_1x^2+c_2\frac{1}{x^2}}=\frac{2(-c_1x^4+c_2)}{x(c_1x^4+c_2)}$$ $$\begin{cases} y=\frac{2(-x^4+C)}{x(x^4+C)}\quad\text{with}\quad c_1\neq 0.\\ y=\frac{2}{x}\quad\text{with}\quad c_1=0.\\ y=-\frac{2}{x}\quad\text{with}\quad c_2=0. \end{cases}$$