Solution of Bernoulli's Equation

Bernoulli's equation is a special type of nonlinear ordinary differential equation (ODE) that can be solved using a specific method. The general form of Bernoulli's equation is:

$$ \frac{dy}{dx} + P(x)y = Q(x)y^n $$

where $n$ is any real number, and $P(x)$ and $Q(x)$ are continuous functions of $x$. When $n = 0$ or $n = 1$, the equation becomes linear and can be solved using methods for linear ODEs. For other values of $n$, the equation is nonlinear.

Transforming Bernoulli's Equation into a Linear ODE

The solution method for Bernoulli's equation involves transforming it into a linear ODE. This is done by making the substitution:

$$ v = y^{1-n} $$

Taking the derivative of $v$ with respect to $x$ gives:

$$ \frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx} $$

Substituting $v$ and $\frac{dv}{dx}$ into the original Bernoulli's equation, we get:

$$ (1-n)y^{-n}\frac{dy}{dx} + (1-n)P(x)y^{1-n} = (1-n)Q(x) $$

Simplifying, we obtain a linear ODE in terms of $v$:

$$ \frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x) $$

This linear ODE can be solved using standard methods for linear ODEs, such as integrating factors.

Solving the Linear ODE

Once we have the linear ODE in terms of $v$, we can solve it using the integrating factor method. The integrating factor, $\mu(x)$, is given by:

$$ \mu(x) = e^{\int (1-n)P(x) dx} $$

Multiplying the linear ODE by $\mu(x)$, we get:

$$ \mu(x)\frac{dv}{dx} + \mu(x)(1-n)P(x)v = \mu(x)(1-n)Q(x) $$

The left-hand side of this equation is the derivative of the product $\mu(x)v$, so we can write:

$$ \frac{d}{dx}[\mu(x)v] = \mu(x)(1-n)Q(x) $$

Integrating both sides with respect to $x$ gives:

$$ \mu(x)v = \int \mu(x)(1-n)Q(x) dx + C $$

where $C$ is the constant of integration. Solving for $v$, we have:

$$ v = \frac{1}{\mu(x)}\left(\int \mu(x)(1-n)Q(x) dx + C\right) $$

Finally, we substitute back $v = y^{1-n}$ to find the solution for $y$:

$$ y^{1-n} = \frac{1}{\mu(x)}\left(\int \mu(x)(1-n)Q(x) dx + C\right) $$

$$ y = \left(\frac{1}{\mu(x)}\left(\int \mu(x)(1-n)Q(x) dx + C\right)\right)^{\frac{1}{1-n}} $$

Examples

Let's consider an example to illustrate the solution process.

Example 1:

Solve the Bernoulli's equation:

$$ \frac{dy}{dx} - \frac{2}{x}y = -\frac{2}{x^3}y^2 $$

Here, $P(x) = -\frac{2}{x}$, $Q(x) = -\frac{2}{x^3}$, and $n = 2$.

1. Make the substitution $v = y^{1-2} = y^{-1}$, so $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$.
2. The equation becomes $\frac{dv}{dx} - \frac{2}{x}v = 2/x^3$.
3. The integrating factor is $\mu(x) = e^{\int -2/x dx} = e^{-2\ln|x|} = x^{-2}$.
4. Multiply through by the integrating factor: $x^{-2}\frac{dv}{dx} - 2x^{-3}v = 2x^{-5}$.
5. Integrate both sides: $\int d(x^{-2}v) = \int 2x^{-5} dx$.
6. Solving the integrals gives $x^{-2}v = -\frac{2}{3}x^{-3} + C$.
7. Solve for $v$: $v = -\frac{2}{3x} + Cx^2$.
8. Substitute back $v = y^{-1}$: $y^{-1} = -\frac{2}{3x} + Cx^2$.
9. Finally, solve for $y$: $y = \frac{1}{-\frac{2}{3x} + Cx^2}$.

Differences and Important Points

Feature	Linear ODE	Bernoulli's Equation
Form	$\frac{dy}{dx} + P(x)y = Q(x)$	$\frac{dy}{dx} + P(x)y = Q(x)y^n$
Linearity	Linear	Nonlinear (unless $n = 0$ or $n = 1$)
Solution Method	Integrating factor, variation of parameters, etc.	Transform to linear ODE, then use linear methods
General Solution	$y = e^{-\int P(x) dx} \left(\int Q(x)e^{\int P(x) dx} dx + C\right)$	$y = \left(\frac{1}{\mu(x)}\left(\int \mu(x)(1-n)Q(x) dx + C\right)\right)^{\frac{1}{1-n}}$

The solution of Bernoulli's equation involves transforming it into a linear ODE, which can then be solved using standard methods for linear ODEs. The substitution $v = y^{1-n}$ is key to this transformation, and the integrating factor method is often used to solve the resulting linear ODE. The final solution for $y$ is obtained by substituting back for $v$ and solving for $y$.