Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.
(x.exp(y/x) + y)dx - x dy = 0

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Identify the given differential equation: $(x \cdot \exp(y/x) + y) \, dx - x \, dy = 0$.

Rewrite the equation in the form $M(x,y) \, dx + N(x,y) \, dy = 0$, where $M(x,y) = x \exp(y/x) + y$ and $N(x,y) = -x$.

Check if the equation is homogeneous by verifying if $M(tx, ty)$ and $N(tx, ty)$ are homogeneous functions of the same degree. Here, substitute $x$ by $tx$ and $y$ by $ty$ in $M$ and $N$ and simplify.

If the equation is homogeneous, introduce the substitution $v = \frac{y}{x}$, which implies $y = vx$ and $dy = v \, dx + x \, dv$.

Rewrite the original differential equation in terms of $v$ and $x$ using the substitution, then separate variables or rearrange to solve the resulting equation.

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Homogeneous Differential Equations

A differential equation is homogeneous if it can be expressed so that each term is a function of the ratio y/x (or x/y). This allows the equation to be simplified by substituting v = y/x, reducing it to a separable form. Recognizing and rewriting the equation in this form is essential for solving it.

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Substitution Method (v = y/x)

The substitution v = y/x transforms the original variables into a single variable, simplifying the differential equation. This substitution converts the equation into one involving v and x, often making it separable and easier to integrate. Understanding how to apply and manipulate this substitution is key to solving homogeneous equations.

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Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. After substitution, the homogeneous equation often becomes separable, enabling integration with respect to each variable independently to find the solution.

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