Question

Solve the Riccati's equaton dy/dx=−1−x2+y2

Answer 1

Answer on Question #69645 – Math – Differential Equations

Question

Solve the Riccati's equation

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

Solution

One searches for the general solution in the following form

$$y(x) = z(x) + y_p(x),$$

where $y_p(x)$ is a particular solution of the given equation.

This substitution transforms the equation into Bernoulli's one of the unknown function $z(x)$.

Let the particular solution be

$$y_p(x) = ax.$$

Substituting it into the differential equation we get

$$\frac{d}{dx}(ax) = -1 - x^2 + (ax)^2$$

or

$$a = -1 + (a^2 - 1)x^2.$$

Value

$$a = -1$$

satisfies the previous equation.

The particular solution is

$$y_p(x) = -x,$$

and the general solution may be found as

$$y(x) = z(x) - x.$$

Substituting it into the Riccati's equation we get

$$\frac{d}{dx}(z(x) - x) = -1 - x^2 + (z(x) - x)^2,$$ $$\frac {d z}{d x} - 1 = - 1 - x ^ {2} + z ^ {2} - 2 x z + x ^ {2},$$ $$\frac {d z}{d x} = z ^ {2} - 2 x z.$$

This is Bernoulli's equation. It can be solved using the substitution $z(x) = u(x)v(x)$ :

$$\frac {d (u v)}{d x} = (u v) ^ {2} - 2 x u v,$$ $$v \frac {d u}{d x} + u \frac {d v}{d x} + 2 x u v - u ^ {2} v ^ {2} = 0,$$ $$\left(\frac {d u}{d x} + 2 x u\right) v + \left(u \frac {d v}{d x} - u ^ {2} v ^ {2}\right) = 0,$$

We get the system

$$\left\{ \begin{array}{l} \frac {d u}{d x} + 2 x u = 0 \\ u \frac {d v}{d x} - u ^ {2} v ^ {2} = 0 \end{array} \right..$$

We need a particular solution of the first equation of the system. Check that it is $u(x) = be^{-x^2}$ :

$$\frac {d \left(b e ^ {- x ^ {2}}\right)}{d x} + 2 b x e ^ {- x ^ {2}} = 0,$$ $$- 2 x b e ^ {- x ^ {2}} + 2 b x e ^ {- x ^ {2}} = 0,$$ $$0 = 0.$$

Let $b = 1$ , so a particular solution is $u(x) = e^{-x^2}$ .

As $u(x) = e^{-x^2} \neq 0$ , the second equation may be reduced to

$$\frac {d v}{d x} - u v ^ {2} = 0$$

or

$$\frac {d v}{d x} - e ^ {- x ^ {2}} v ^ {2} = 0.$$

This is the separable DE.

$$\int \frac {d v}{v ^ {2}} = \int e ^ {- x ^ {2}} d x,$$ $$- \frac {1}{v (x)} = \frac {\sqrt {\pi}}{2} \operatorname {erf} x - \frac {C}{2},$$

where $C$ is an integration constant.

Then

$$v (x) = - \frac {1}{\frac {\sqrt {\pi}}{2} \operatorname {erf} x - \frac {C}{2}} = \frac {2}{C - \sqrt {\pi} \operatorname {erf} x}$$

and

$$z (x) = u (x) v (x) = \frac {2 e ^ {- x ^ {2}}}{C - \sqrt {\pi} \operatorname {erf} x}.$$

Finally, the general solution of the Riccati's equation is

$$y (x) = z (x) - x = \frac {2 e ^ {- x ^ {2}}}{C - \sqrt {\pi} \operatorname {erf} x} - x.$$

Answer: $y(x) = \frac{2e^{-x^2}}{C - \sqrt{\pi}\operatorname{erf}x} - x.$

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