Riccati differerntial equation

Question

My book suggest to use the substitution $y(t)=y_{1}(t)+u(t)$ for Riccati odes

Given the ode:

$$y'=1+x^2-y^2$$

$\frac{dy}{dx}=1+x^2-y^2$

A solution is $y_{1}=x$ so $y=x+u$

$\frac{dy}{dx}=1+\frac{du}{dx}$

$1+\frac{du}{dx}=1+x^2-y^2$

$\frac{du}{dx}=x^2-(x+u)^2$

$\frac{du}{dx}=x^2-x^2-2xu-u^2$

$\frac{du}{dx}=-2xu-u^2$

How do I procceed from here? Is the substitution incorrect?

Answer 1

$$y'=1+x^2-y^2$$ $$y_1'=1+x^2-y_1^2$$

the difference gives

$$(y-x)'=(x-y)(x+y)=(x-y)(y-x+2x)$$

$$u'=-u(u+2x)=-2xu-u^2$$

$$-\frac{u'}{u^2}=2x\frac{1}{u}+1$$

this is Bernouilli type.

put $z=\frac{1}{u}$.

thus $$z'=2xz+1$$

Answer 2

Set $u = \frac{v'}{v}$ for some $v$ yet to be determined, then we see that \begin{align} u' = \frac{v''v-(v')^2}{v^2} \end{align} then we see that \begin{align} u'=-2xu-u^2 \ \ \implies&\ \ \frac{v''v-v'^2}{v^2}=-2x\frac{v'}{v}-\frac{v'^2}{v^2}\\ \implies& \ \ \frac{v''}{v}=-2x\frac{v'}{v}\\ \implies& \ \ v''=-2xv'. \end{align}

Answer 3

Yes you are correct $$u'=-2xu-u^2$$ This is Bernouilli's equation. Divide by $u^2$ $$\frac {u'}{u^2}=-2x\frac 1u-1$$ Substitute $z=1/u$ $$z'-2xz=1$$ This is a linear first order equation

$$(ze^{-x^2})'=e^{-x^2}$$ $$u^{-1}(x)=ke^{x^2}+e^{x^2}\int e^{-x^2}dx$$

The integral can be expressed with the error function