To show that the function [tex]\( u(x, y) = x \sqrt{x y} \)[/tex] satisfies the equation [tex]\( x u_x - y u_y = u \)[/tex] subject to [tex]\( u(y, y) = y^2 \)[/tex] for [tex]\( y \geq 0 \)[/tex], we will go through several steps.
### Step 1: Compute the partial derivatives of [tex]\( u \)[/tex]
First, let's find the partial derivatives of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. Partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ u_x = \frac{\partial}{\partial x} \left( x \sqrt{x y} \right) \][/tex]
To differentiate, we use the product rule:
[tex]\[ u_x = \frac{\partial}{\partial x} \left( x (xy)^{1/2} \right) \][/tex]
Let [tex]\( f(x) = x \)[/tex] and [tex]\( g(x) = (xy)^{1/2} \)[/tex]:
[tex]\[ u_x = f'(x) g(x) + f(x) g'(x) \][/tex]
[tex]\[ f'(x) = 1 \quad \text{and} \quad g'(x) = \frac{1}{2} (xy)^{-1/2} \cdot y = \frac{y}{2 \sqrt{xy}} \][/tex]
[tex]\[ u_x = 1 \cdot (xy)^{1/2} + x \cdot \frac{y}{2 \sqrt{xy}} \][/tex]
[tex]\[ = \sqrt{xy} + \frac{x y}{2 \sqrt{xy}} \][/tex]
[tex]\[ = \sqrt{xy} + \frac{x y}{2 \sqrt{xy}} = \sqrt{xy} + \frac{1}{2} \sqrt{xy} \][/tex]
[tex]\[ = \frac{3}{2} \sqrt{xy} \][/tex]
2. Partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ u_y = \frac{\partial}{\partial y} \left( x \sqrt{x y} \right) \][/tex]
To differentiate, we use the chain rule:
[tex]\[ u_y = x \cdot \frac{\partial}{\partial y} \left( (xy)^{1/2} \right) \][/tex]
[tex]\[ = x \cdot \frac{1}{2} (xy)^{-1/2} \cdot x \][/tex]
[tex]\[ = x \cdot \frac{x}{2 \sqrt{xy}} \][/tex]
[tex]\[ = \frac{x^2}{2 \sqrt{xy}} \][/tex]
[tex]\[ = \frac{x \sqrt{xy}}{2y} \][/tex]
Thus, we have:
[tex]\[ u_x = \frac{3 \sqrt{xy}}{2} \][/tex]
[tex]\[ u_y = \frac{x \sqrt{xy}}{2y} \][/tex]
### Step 2: Verify the given equation [tex]\( x u_x - y u_y = u \)[/tex]
Next, we need to plug these partial derivatives into [tex]\( x u_x - y u_y \)[/tex] and verify if it equals [tex]\( u \)[/tex].
Substitute [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex]:
[tex]\[ x u_x - y u_y = x \left(\frac{3 \sqrt{xy}}{2}\right) - y \left(\frac{x \sqrt{xy}}{2y}\right) \][/tex]
[tex]\[ = \frac{3 x \sqrt{xy}}{2} - \frac{x \sqrt{xy}}{2} \][/tex]
[tex]\[ = \frac{3 x \sqrt{xy} - x \sqrt{xy}}{2} \][/tex]
[tex]\[ = \frac{2 x \sqrt{xy}}{2} \][/tex]
[tex]\[ = x \sqrt{xy} \][/tex]
Since [tex]\( u(x, y) = x \sqrt{x y} \)[/tex], we have shown:
[tex]\[ x u_x - y u_y = u \][/tex]
### Step 3: Check the specific case [tex]\( u(y, y) = y^2 \)[/tex] for [tex]\( y \geq 0 \)[/tex]
Now, we need to verify the condition [tex]\( u(y, y) = y^2 \)[/tex].
Substitute [tex]\( x = y \)[/tex] into [tex]\( u \)[/tex]:
[tex]\[ u(y, y) = y \sqrt{y \cdot y} \][/tex]
[tex]\[ = y \sqrt{y^2} \][/tex]
[tex]\[ = y \cdot y \][/tex]
[tex]\[ = y^2 \][/tex]
As [tex]\( u(y, y) = y^2 \)[/tex] is indeed satisfied for [tex]\( y \geq 0 \)[/tex], the given condition holds.
### Conclusion
We have verified that [tex]\( u(x, y) = x \sqrt{x y} \)[/tex] satisfies both the differential equation [tex]\( x u_x - y u_y = u \)[/tex] and the condition [tex]\( u(y, y) = y^2 \)[/tex] for [tex]\( y \geq 0 \)[/tex].
