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# Homogeneous first order ODE

A homogeneous first-order ordinary differential equation (ODE) is an equation that can be written in the form $$y\prime = f(x, y) = g(\frac{y}{x})$$, where g is a function of the ratio $$\frac{y}{x}$$. In other words, if we substitute $$v = \frac{y}{x}$$, the equation becomes $$y\prime = g(v)$$, which is separable and can be solved using integration.

Alternatively, we can rewrite the homogeneous first-order ODE as $$y\prime = F(\frac{y}{x})$$, where $$F$$ is some function of the ratio $$\frac{y}{x}$$. In this form, we can make a substitution $$u = \frac{y}{x}$$, so that $$y\prime = u + xu\prime$$, and the equation becomes a separable first-order ODE in $$u$$ and $$x$$.

Solving a homogeneous first-order ODE involves finding a function $$y(x)$$ that satisfies the equation. The general solution to a homogeneous first-order ODE involves one arbitrary constant, which can be determined by applying initial or boundary conditions.

## Degree of the function

In order to understand the terminology, used in this lecture, I need to explain briefly what is the degree of function.

If we have function $$F(x, y)$$ and we can write the following equality:

$f(t\times x, t\times y) = t^n\times F(x, y)$

Then we can say, that function $$F(x, y)$$ have a degree of $$n$$, where $$n$$ can be any integer number, including 0.

### Examples of calculating the degree of a function

To clarify this rule, let's consider a few examples:

Example 1: $$xy + x^2$$

$F(x, y) = xy + x^2 \\ F(tx, ty) = txty + (tx)^2 = t^2(xy + x^2) \Rightarrow \boxed{n = 2}$

Example 2: $$\sin{(\frac{x}{y})}$$

$F(x, y) = \sin{(\frac{x}{y})} \\ F(tx, ty) = \sin{(\frac{tx}{ty})} = t^0\times\sin{(\frac{x}{y})} \Rightarrow \boxed{n = 0}$

Example 3: $$\sin{x} + \cos{y}$$

$F(x, y) = \sin{x} + \cos{y} \\ F(tx, ty) = \sin{tx} + \cos{ty} = \cdots \Rightarrow \boxed{Non-homogeniuos}$

## How to solve homogeneous differential equation

If we have an equation in the form of $$M(x, y)\frac{\mathop{dy}}{\mathop{dx}} = N(x, y)$$, when the functions $$M(x, y)$$ and $$N(x, y)$$ have the same degree, then we can introduce function $$F(x, y) = \frac{M(x, y)}{N(x, y)}$$ with the power of $$0$$ and rewrite it as $$F(x, y) = Q(\frac{y}{x})$$. Then we can represent our homogeneous differential equation as:

$\boxed{\frac{\mathop{dy}}{\mathop{dx}} = Q(\frac{y}{x})}$

To solve it we need to introduce substitution $$v = \frac{y}{x} \Rightarrow y = vx$$ and therefore, let's do a few transformations:

$\frac{\mathop{dy}}{\mathop{dx}} = \frac{\mathop{d(vx)}}{\mathop{dx}} = \frac{\mathop{dv}}{\mathop{dx}}x + v \frac{\mathop{dx}}{\mathop{dx}} = \frac{\mathop{dv}}{\mathop{dx}}x + v \\ Q(v) = Q(\frac{y}{x}) = \frac{\mathop{dy}}{\mathop{dx}} = v + x\frac{\mathop{dv}}{\mathop{dx}} \\ x\frac{\mathop{dv}}{\mathop{dx}} = Q(v) - v \\ \frac{\mathop{dv}}{Q{v} - v} = \frac{\mathop{dx}}{x} \\ \boxed{\int\frac{\mathop{dv}}{Q{v} - v} = \int\frac{\mathop{dx}}{x}}$

That it! Solved. Later I will expain it in more details on some particular examples.

For more details about Homogeneous first order ODE, please watch this video:

Published: 2023-05-14 13:39:30
Updated: 2023-05-14 15:34:30

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