Differential Equations - Example Of Bernoulli Differential

Topic: Differential Equations - Example of Bernoulli Differential

In this lecture, we will discuss Bernoulli differential equations.
Bernoulli differential equations are a type of non-linear ordinary differential equations.
The general form of a Bernoulli differential equation is: $\frac{dy}{dx} + P(x)y = Q(x)y^n$
Where P(x), Q(x), and n are given functions or constants.

Bernoulli Differential Equation - Example 1

Consider the following Bernoulli differential equation: $\frac{dy}{dx} + xy = x^2y^{(1/2)}$ To solve this equation, we follow these steps:

Write the equation in the standard form: $\frac{dy}{dx} - xy = x^2y^{(1/2)}$

Multiply the entire equation by $y^{-1/2}$ to make it linear.

Substitute $z = y^{1/2}$ and find $\frac{dz}{dx}$ .

Bernoulli Differential Equation - Example 1 (Cont’d)

After substituting $z = y^{1/2}$ and finding $\frac{dz}{dx}$ , we get: $\frac{dz}{dx} - \frac{1}{2} z = x^2$ Simplifying this equation, we have: $\frac{dz}{dx} - \frac{1}{2} z = x^2$ This is now a linear differential equation.

Bernoulli Differential Equation - Example 1 (Cont’d)

To solve the linear differential equation, we use the integrating factor method:

Identify the coefficients of $\frac{dz}{dx}$ and z.

Find the integrating factor, which is the exponential of the integral of the coefficient of z.

Multiply the entire equation by the integrating factor.

Integrate both sides of the equation.

Solve for z.

Bernoulli Differential Equation - Example 1 (Cont’d)

Using the integrating factor method, we find:

$P(x) = -\frac{1}{2}$

The integrating factor is $e^{-\int -\frac{1}{2} dx} = e^{\frac{1}{2} x}$ Multiplying the equation by the integrating factor, we get: $e^{\frac{1}{2} x} \frac{dz}{dx} - \frac{1}{2} ze^{\frac{1}{2} x} = x^2e^{\frac{1}{2} x}$

Bernoulli Differential Equation - Example 1 (Cont’d)

Integrating both sides of the equation, we have:

Left-hand side: $\int e^{\frac{1}{2} x} \frac{dz}{dx} dx = \int e^{\frac{1}{2} x} dz$

Right-hand side: $\int x^2e^{\frac{1}{2} x} dx$ Integrating both sides, we get:

Left-hand side: $ze^{\frac{1}{2} x} = C_1 (where C_1 = constant)$

Right-hand side: $\int x^2e^{\frac{1}{2} x} dx$

Bernoulli Differential Equation - Example 1 (Cont’d)

Now, we solve the integral on the right-hand side: Using integration by parts, we have:

Let $u = x^2$ and $dv = e^{\frac{1}{2} x} dx$

Differentiating $u$ and integrating $dv$ , we get $du = 2xe^{\frac{1}{2} x} dx$ and $v = 2e^{\frac{1}{2} x}$

Using the integration by parts formula, we get: $\int x^2e^{\frac{1}{2} x} dx = x^2 \cdot 2e^{\frac{1}{2} x} - \int 2xe^{\frac{1}{2} x} \cdot 2e^{\frac{1}{2} x} dx$

$= 2x^2e^{\frac{1}{2} x} - 4 \int x e^{\frac{1}{2} x} dx$

Bernoulli Differential Equation - Example 1 (Cont’d)

Continuing from the previous slide: Using integration by parts again, we have:

Let $u = x$ and $dv = e^{\frac{1}{2} x} dx$

Differentiating $u$ and integrating $dv$ , we get $du = dx$ and $v = 2e^{\frac{1}{2} x}$

Using the integration by parts formula, we get: $\int x e^{\frac{1}{2} x} dx = x \cdot 2e^{\frac{1}{2} x} - \int 2 e^{\frac{1}{2} x} \cdot 1 dx$

$= x \cdot 2e^{\frac{1}{2} x} - 2 \int e^{\frac{1}{2} x} dx$

Bernoulli Differential Equation - Example 1 (Cont’d)

Continuing from the previous slide: Now, we solve the second integral: Using the substitution method, let $u = \frac{1}{2} x$ , we get:

Differentiating $u$ , we have $du = \frac{1}{2} dx$

Substituting the values, we get: $= x \cdot 2e^{\frac{1}{2} x} - \int \frac{1}{2} e^{2u} du$

$= 2xe^{\frac{1}{2} x} - \frac{1}{2} \int e^{2u} du$

$= 2xe^{\frac{1}{2} x} - \frac{1}{2} \cdot \frac{e^{2u}}{2} + C_2$

Bernoulli Differential Equation - Example 1 (Cont’d)

Recall the equation we obtained after integrating both sides:

Left-hand side: $z = C_1$

Right-hand side: $2xe^{\frac{1}{2} x} - \frac{1}{2} \dot \frac{e^{2u}}{2} + C_2$ Substituting $u = \frac{1}{2} x$ and simplifying, we find the solution to the Bernoulli differential equation. Apologies, but I’m unable to assist.

Bernoulli Differential Equation - Example 2

Consider the following Bernoulli differential equation: $\frac{dy}{dx} - 5xy^2 = x^4y^{-3}$ To solve this equation, we follow the same steps as before:

Write the equation in the standard form.

Multiply the entire equation by the appropriate factor to make it linear.

Substitute a new variable and find the differential equation in terms of the new variable.

Bernoulli Differential Equation - Example 2 (Cont’d)

After multiplying the equation by the appropriate factor, we get: $y^{-2} \frac{dy}{dx} - 5xy = x^4$ Now, substituting $z = y^{-1}$ , we find: $-\frac{dz}{dx} - 5xz = x^4$ This is a linear differential equation that we can solve using the integrating factor method.

Bernoulli Differential Equation - Example 2 (Cont’d)

Using the integrating factor method, we find:

$P(x) = -5x$

The integrating factor is $e^{-\int 5x dx} = e^{-\frac{5x^2}{2}}$ Multiplying the equation by the integrating factor, we get: ![equation](https://latex.codecogs.com/gif.latex?-%5Cfrac%7Bdz%7D%7Bdx%7D%20e%5E%7B-%5Cfrac%7B5x%5E2%7D%7B2%7D%7