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Homogeneous first order ODE
Homogeneous first order ODEA homogeneous firstorder 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 firstorder 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 firstorder ODE in \(u\) and \(x\). Solving a homogeneous firstorder ODE involves finding a function \(y(x)\) that satisfies the equation. The general solution to a homogeneous firstorder ODE involves one arbitrary constant, which can be determined by applying initial or boundary conditions. Degree of the functionIn 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 functionTo 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: \(\sin{x} + \cos{y}\) \[ F(x, y) = \sin{x} + \cos{y} \\ F(tx, ty) = \sin{tx} + \cos{ty} = \cdots \Rightarrow \boxed{Nonhomogeniuos}\]How to solve homogeneous differential equationIf 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:

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