How to solve homogeneous differential equations by substitution

📚 Understanding Homogeneous Differential Equations

A homogeneous differential equation is one where, informally, if you scale the variables, the equation remains the same. More precisely, a first-order differential equation of the form $\frac{dy}{dx} = f(x, y)$ is homogeneous if the function $f(x, y)$ is a homogeneous function of degree zero. This means that for any scalar $t$, $f(tx, ty) = f(x, y)$.

📜 History and Background

The concept of homogeneous differential equations and their solution methods have roots in the development of calculus and differential equations by mathematicians such as Leibniz and Bernoulli in the 17th and 18th centuries. The method of substitution provides a way to transform these equations into a separable form, making them solvable using standard integration techniques.

🔑 Key Principles of Solving by Substitution

The core idea is to introduce a new variable, typically $v = \frac{y}{x}$ (or sometimes $v = \frac{x}{y}$, depending on the equation's structure), which allows us to rewrite the original equation in terms of $v$ and $x$ (or $y$), hopefully leading to a separable equation.

• 🔄 Substitution: Let $v = \frac{y}{x}$, so $y = vx$. Then, $\frac{dy}{dx} = v + x\frac{dv}{dx}$. This is the crucial step.
• 📝 Rewrite the Equation: Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the original differential equation.
• 🧩 Simplify and Separate: Algebraically manipulate the equation to separate variables, getting terms involving $v$ on one side and terms involving $x$ on the other. You should aim for an equation of the form $g(v) dv = h(x) dx$.
• 🧪 Integrate: Integrate both sides of the separated equation with respect to their respective variables: $\int g(v) dv = \int h(x) dx$.
• 🔙 Back-Substitute: Replace $v$ with $\frac{y}{x}$ in the solution to obtain the solution in terms of $x$ and $y$.

🌍 Real-world Examples

Let's walk through some examples to solidify the concept:

Example 1: Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$

1. Let $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
2. Substitute: $v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{x(vx)} = \frac{x^2 + v^2x^2}{vx^2} = \frac{1 + v^2}{v}$.
3. Simplify: $x\frac{dv}{dx} = \frac{1 + v^2}{v} - v = \frac{1}{v}$.
4. Separate: $v dv = \frac{dx}{x}$.
5. Integrate: $\int v dv = \int \frac{dx}{x} \implies \frac{1}{2}v^2 = \ln|x| + C$.
6. Back-substitute: $\frac{1}{2}(\frac{y}{x})^2 = \ln|x| + C \implies y^2 = 2x^2(\ln|x| + C)$.

Example 2: Solve $\frac{dy}{dx} = \frac{y}{x} + e^{\frac{y}{x}}$

1. Let $v = \frac{y}{x}$. Then $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
2. Substitute: $v + x\frac{dv}{dx} = v + e^v$.
3. Simplify: $x\frac{dv}{dx} = e^v$.
4. Separate: $e^{-v} dv = \frac{dx}{x}$.
5. Integrate: $\int e^{-v} dv = \int \frac{dx}{x} \implies -e^{-v} = \ln|x| + C$.
6. Back-substitute: $-e^{-\frac{y}{x}} = \ln|x| + C \implies e^{-\frac{y}{x}} = -\ln|x| - C$.

📝 Conclusion

Solving homogeneous differential equations by substitution is a powerful technique that transforms a seemingly complex problem into a manageable, separable equation. By understanding the underlying principles and practicing with examples, you can master this method and confidently tackle a wide range of differential equations.