We are tasked with proving that the substitution $y = vx$ can help solve any equation of the form:

$y^n f(x) dx + H(x, y)(y dx - x dy) = 0$

where $H(x, y)$ is a homogeneous function of $y$ and $x$. Let's proceed step by step.

Step 1: Apply the substitution $y = vx$

Given the substitution $y = vx$, the first step is to find the derivatives of $y$ with respect to $x$.

Since $y = vx$, differentiating both sides with respect to $x$, we get:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

This expression will be useful in simplifying the equation after the substitution.

Step 2: Rewrite the equation using $y = vx$

Substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the original equation:

$y^n f(x) dx + H(x, y)(y dx - x dy) = 0$

We replace all occurrences of $y$ with $vx$, so:

• $y^n = (vx)^n = v^n x^n$,
• $dy = v dx + x dv$.

Thus, the equation becomes:

$(v^n x^n) f(x) dx + H(x, vx) \left( vx dx - x(v dx + x dv) \right) = 0$

Now simplify the second term:

$vx dx - x(v dx + x dv) = vx dx - (xv dx + x^2 dv) = -x^2 dv$

Thus, the equation becomes:

$v^n x^n f(x) dx + H(x, vx)(-x^2 dv) = 0$

Step 3: Simplify the equation

Divide the entire equation by $x^n$ to eliminate powers of $x$:

$v^n f(x) dx - H(x, vx) x^{2-n} dv = 0$

This equation is now separable in terms of $x$ and $v$. Notice that the homogeneous function $H(x, vx)$ can be written as a function of $v$ only, because $H(x, vx)$ is homogeneous. Let $H(x, vx) = \tilde{H}(v)$, where $\tilde{H}(v)$ is a function of $v$ alone.

Thus, the equation simplifies further to:

$v^n f(x) dx - \tilde{H}(v) x^{2-n} dv = 0$

Step 4: Solve the equation

Now, we can separate variables $x$ and $v$:

$\frac{f(x)}{x^{2-n}} dx = \frac{\tilde{H}(v)}{v^n} dv$

This equation can be integrated on both sides, since the variables $x$ and $v$ are completely separated.

Let:

$\int \frac{f(x)}{x^{2-n}} dx = \int \frac{\tilde{H}(v)}{v^n} dv$

After performing the integrals, we will arrive at a general solution that relates $x$ and $v$. Since $v = \frac{y}{x}$, this provides a way to express the solution in terms of $x$ and $y$.

Conclusion:

The substitution $y = vx$ allows us to reduce the original differential equation to a separable form, where $x$ and $v$ (which is related to $y$) can be separated and integrated. This proves that the substitution $y = vx$ can be used to solve any equation of the given form, where $H(x, y)$ is a homogeneous function.

Would you like further details on any step or a more specific example?

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Here are 5 related questions you may find helpful:

1. What are homogeneous functions, and how do they help in solving differential equations?
2. How can we identify whether a differential equation is homogeneous?
3. Can the substitution $y = vx$ be used for non-homogeneous equations?
4. How do we solve homogeneous equations in polar coordinates?
5. What are some other common substitutions for solving differential equations?

Tip: When dealing with homogeneous functions, check if scaling the variables $x$ and $y$ by the same factor results in a proportional change to the function. This property is key to simplifying many differential equations.