To demonstrate that the function [tex]\( u(x, y) = x \sqrt{x y} \)[/tex] satisfies the equation [tex]\( x u_x - y u_y = u \)[/tex], we need to perform the following steps. Let's do this step-by-step:
1. Find the partial derivatives [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex]:
- Partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[
u(x, y) = x \sqrt{x y}
\][/tex]
To find [tex]\( u_x \)[/tex], use the product rule and the chain rule:
[tex]\[
u_x = \frac{\partial}{\partial x} (x \cdot \sqrt{x y})
\][/tex]
First, recognize that [tex]\( \sqrt{x y} = (x y)^{1/2} \)[/tex].
Then,
[tex]\[
u(x, y) = x \cdot (x y)^{1/2}
\][/tex]
Now, apply the product rule:
[tex]\[
u_x = \frac{\partial x}{\partial x} \cdot (x y)^{1/2} + x \cdot \frac{\partial (x y)^{1/2}}{\partial x}
\][/tex]
We know [tex]\( \frac{\partial x}{\partial x} = 1 \)[/tex]:
[tex]\[
u_x = (x y)^{1/2} + x \cdot \frac{\partial (x y)^{1/2}}{\partial x}
\][/tex]
Using the chain rule for [tex]\( (x y)^{1/2} \)[/tex]:
[tex]\[
\frac{\partial (x y)^{1/2}}{\partial x} = \frac{1}{2} (x y)^{-1/2} \cdot y = \frac{y}{2 \sqrt{x y}}
\][/tex]
Thus,
[tex]\[
u_x = \sqrt{x y} + x \cdot \frac{y}{2 \sqrt{x y}} = \sqrt{x y} + \frac{x y}{2 \sqrt{x y}} = \sqrt{x y} + \frac{1}{2} \sqrt{x y} = \frac{3}{2} \sqrt{x y}
\][/tex]
2. Partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[
u(x, y) = x \sqrt{x y}
\][/tex]
To find [tex]\( u_y \)[/tex]:
[tex]\[
u_y = \frac{\partial}{\partial y} (x \cdot \sqrt{x y})
\][/tex]
Recognize [tex]\( \sqrt{x y} = (x y)^{1/2} \)[/tex]:
[tex]\[
u_y = x \cdot \frac{\partial (x y)^{1/2}}{\partial y}
\][/tex]
Using the chain rule for [tex]\( (x y)^{1/2} \)[/tex]:
[tex]\[
\frac{\partial (x y)^{1/2}}{\partial y} = \frac{1}{2} (x y)^{-1/2} \cdot x = \frac{x}{2 \sqrt{x y}}
\][/tex]
Thus,
[tex]\[
u_y = x \cdot \frac{x}{2 \sqrt{x y}} = \frac{x^2}{2 \sqrt{x y}}
\][/tex]
Simplify further:
[tex]\[
u_y = \frac{x^2}{2 \sqrt{x y}}
\][/tex]
3. Verify [tex]\( x u_x - y u_y = u \)[/tex]:
- Substitute [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex] into the equation:
[tex]\[
x u_x - y u_y = x \left( \frac{3}{2} \sqrt{x y} \right) - y \left( \frac{x^2}{2 \sqrt{x y}} \right)
\][/tex]
Simplify each term:
[tex]\[
x u_x = x \cdot \frac{3}{2} \sqrt{x y} = \frac{3}{2} x \sqrt{x y}
\][/tex]
[tex]\[
y u_y = y \cdot \frac{x^2}{2 \sqrt{x y}} = \frac{1}{2} y \cdot \frac{x^2}{\sqrt{x y}} = \frac{1}{2} \frac{yx^2}{\sqrt{x y}} = \frac{1}{2} x^2 \sqrt{\frac{y}{x}} y = \frac{1}{2} x \sqrt{x y}
\][/tex]
Therefore,
[tex]\[
x u_x - y u_y = \frac{3}{2} x \sqrt{x y} - \frac{1}{2} x \sqrt{x y} = \sqrt{x y}
\][/tex]
Recognize that
[tex]\[
u = x \sqrt{x y}
\][/tex]
Hence,
[tex]\(
x u_x - y u_y = u
\)[/tex]
Thus, we have successfully shown that [tex]\( x u_x - y u_y = u \)[/tex].
Lastly, let's check the condition [tex]\( u(y, y) = y^2 \)[/tex]:
[tex]\[
u(y, y) = y \sqrt{y \cdot y} = y \sqrt{y^2} = y \cdot y = y^2
\][/tex]
Therefore, the function [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].
