Bernoulli, Riccati Equation

If a differential equation is written in the form

$$y^{\prime} + P(x)y = Q(x)y^{\alpha} \ (\alpha \neq 0, 1),$$

then it is called the Bernoulli's equation. we first get rid of $y^{\alpha}$ by multiplying $y^{-\alpha}$ to both sides of equation.

$$y^{-\alpha}y^{\prime} + P(x)y^{(1 - \alpha)} = Q(x)$$

Now let

$$u = y^{(1 - \alpha)}$$

Then $u = y^{(1 - \alpha)}$, $u^{\prime} = (1 - \alpha)y^{-\alpha}y^{\prime}$ implies

$$\frac{1}{(1 - \alpha)}u^{\prime} + P(x)u = Q(x).$$

Note that this is a 1st order linear.

Example 1..18   Find the general solution of $y^{\prime} + \frac{1}{x}y = -x^{3}y^{2}$.

SOLUTION This is a Bernoulli's equation. Then multiply $y^{-2}$ to both sides of equation.

$$y^{-2}y^{\prime} + \frac{1}{x}y^{-1} = -x^{3}$$

Now let $u = y^{-1}$. Then $u^{\prime} = -y^{-2}y^{\prime}$ and

$$-u^{\prime} + \frac{1}{x}u = -x^{3}$$

This is a linear differential equation in $u$. So, rewrite this into the normal form.

$$u^{\prime} - \frac{1}{x}u = x^{3}$$

Then $\mu = e^{-\int 1/x dx} = e^{-\log{x}} = 1/x$. Multiplying $\mu$ to both sides of the equation and noting the left-hand side becomes the derivative of the integrating factor times the independent variable $u$. Thus we have

$$d(\frac{u}{x}) = x^{2}dt$$

Integrating both sides to obtain the general solution.

$$u = y^{-1} = \frac{x^{4}}{3} + cx \ensuremath{\ \blacksquare}$$

Given

$$\frac{df(y)}{dy}\frac{dy}{dx} + P(x)f(y) = Q(x),$$

let $u = f(y)$. Then since $\frac{du}{dx} = \frac{df(y)}{dy}\frac{dy}{dx}$,

$$\frac{du}{dx} + P(x)u = Q(x)$$

Example 1..19   Find the general solution of $\cos{y}\frac{dy}{dx} + \frac{1}{x}\sin{y} = 1$.

SOLUTION Let $u = \sin{y}$. Then $\frac{du}{dx} = \cos{y}\frac{dy}{dx}$. Thus the given differential equation is expressed in the form

$$\frac{du}{dx} + \frac{1}{x}u = 1.$$

This is a linear differential equation in $u$. Then the integrating factor $\mu = \int e^{\frac{1}{x}}dx = e^{\log{x}} = x$. Now multiply $\mu$ to both sides of the equation to get

$$d(xu) = xdx$$

Then $xu = \frac{1}{2}x^2 + c$. Solve for $u$ to get

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

Since $u = \sin{y}$,

$$\sin{y} = \frac{1}{2}x + cx^{-1}\ensuremath{\ \blacksquare}$$

The differential equation expressed in the form

$$\frac{dy}{dx} = A(x)y^{2} + B(x)y + C(x)$$

is called the Riccati's equation. If a solution $f(x)$ is found, then we let $y = f(x) + \frac{1}{u}$. Now $\frac{dy}{dx} = f'(x) - \frac{1}{u^2}\frac{du}{dx}$. Thus

$$f'(x) - \frac{1}{u^2}\frac{du}{dx} = A(x)(f(x) + \frac{1}{u})^2 + B(x)(f(x) + \frac{1}{u}) + C(x)$$

This is a linear differential equation in $u$.

Example 1..20   Solve the following differential equation provided a solution $f(x) = 1$.

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

SOLUTION This is a Riccati's equation. Since $f(x) = 1$ is a solution to the equation. Thus let

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

Then $y^{\prime} = -u^{\prime}/u^{2}$. Substitute this into the differetial equation.

$$-\frac{u^{\prime}}{u^{2}} = (1-x)(1 + \frac{1}{u})^{2} + (2x-1)( 1 + \frac{1}{u}) - x .$$

Now simplify the equation. Then we have

$$u^{\prime} + u = 1 - x$$

Multiply $\mu = e^{\int dx} = e^{x}$ to both sides of the equation to get

$$d(e^{x}u) = e^{x}(1-x)dx .$$

Integrating both sides,

$$e^x u = e^x - xe^x + e^x + c = 2e^x - xe^x + c$$

Simplifying to get

$$u = 2 - x + ce^{-x} .$$

Since $y = 1 + 1/u$, we have

$$y = 1 + \frac{1}{2 - x + ce^{-x}}\ensuremath{\ \blacksquare}$$