Solving the Homogeneous Differential Equation

The given differential equation is $x dy - y dx = \sqrt{x^2 + y^2} dx$, with the initial condition $y(1) = 0$ and $x > 0$. We need to find the value of $y(3)$.

First, rewrite the equation in the standard form $\frac{dy}{dx}$:

$x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}$

$x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}$

$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$

This is a homogeneous differential equation because the right-hand side can be expressed as a function of $y/x$. Let $v = \frac{y}{x}$, which implies $y = vx$. Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting and Separating Variables

Substitute $v$ and $\frac{dy}{dx}$ into the equation:

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + (vx)^2}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2(1 + v^2)}}{x}$

Since $x > 0$, $\sqrt{x^2} = x$.

$v + x \frac{dv}{dx} = \frac{vx + x\sqrt{1 + v^2}}{x}$

$v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$

$x \frac{dv}{dx} = \sqrt{1 + v^2}$

Separate the variables:

$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$

Integrating the Equation

Integrate both sides:

$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$

The standard integral $\int \frac{1}{\sqrt{1+v^2}} dv$ is $\ln(v + \sqrt{1+v^2})$. The integral $\int \frac{1}{x} dx$ is $\ln|x|$. Since $x > 0$, this is $\ln(x)$.

$\ln(v + \sqrt{1 + v^2}) = \ln(x) + C$

Where $C$ is the constant of integration.

Applying the Initial Condition

Use the initial condition $y(1) = 0$. When $x=1$, $y=0$, so $v = \frac{y}{x} = \frac{0}{1} = 0$. Substitute $x=1$ and $v=0$ into the integrated equation:

$\ln(0 + \sqrt{1 + 0^2}) = \ln(1) + C$

$\ln(1) = 0 + C$

$0 = C$

The equation becomes:

$\ln(v + \sqrt{1 + v^2}) = \ln(x)$

Finding the Solution Curve y(x)

Exponentiate both sides:

$v + \sqrt{1 + v^2} = x$

Substitute back $v = \frac{y}{x}$:

$\frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = x$

$\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = x$

$y + \sqrt{x^2 + y^2} = x^2$

$\sqrt{x^2 + y^2} = x^2 - y$

Square both sides (note: requires $x^2 - y \ge 0$):

$x^2 + y^2 = (x^2 - y)^2$

$x^2 + y^2 = x^4 - 2x^2y + y^2$

$x^2 = x^4 - 2x^2y$

Rearrange to solve for $y$:

$2x^2y = x^4 - x^2$

Since $x > 0$, $x^2 \neq 0$, we can divide by $2x^2$:

$y = \frac{x^4 - x^2}{2x^2}$

$y = \frac{x^2(x^2 - 1)}{2x^2}$

$y = \frac{x^2 - 1}{2}$

The condition $x^2 - y \ge 0$ becomes $x^2 - \frac{x^2 - 1}{2} = \frac{2x^2 - x^2 + 1}{2} = \frac{x^2+1}{2}$, which is always positive for $x>0$. The solution is valid.

Calculating y(3)

Substitute $x=3$ into the solution $y = \frac{x^2 - 1}{2}$:

$y(3) = \frac{3^2 - 1}{2}$

$y(3) = \frac{9 - 1}{2}$

$y(3) = \frac{8}{2}$

$y(3) = 4$

The value of $y(3)$ is 4.