1. **Problem Statement:** Check if each function $u(x,y)$ is homogeneous and find its degree if yes. Then compute: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$$ and $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}$$ for each function. --- 2. **Function (i):** $$u(x,y) = -\frac{1}{x^2} + \frac{1}{xy} + \frac{\log x - \log y}{x^2 + y^2}$$ - Check homogeneity by replacing $(x,y)$ with $(tx, ty)$: $$u(tx,ty) = -\frac{1}{t^2 x^2} + \frac{1}{t^2 xy} + \frac{\log(tx) - \log(ty)}{t^2 (x^2 + y^2)} = \frac{1}{t^2} \left(-\frac{1}{x^2} + \frac{1}{xy} + \frac{\log x - \log y}{x^2 + y^2}\right)$$ - Note $\log(tx) - \log(ty) = \log t + \log x - (\log t + \log y) = \log x - \log y$ so numerator unchanged. - So $u(tx,ty) = t^{-2} u(x,y)$, function is homogeneous of degree $-2$. - Compute first derivatives: $$\frac{\partial u}{\partial x} = \frac{2}{x^3} - \frac{1}{x^2 y} + \text{(complex term from last fraction)}$$ - For brevity, use Euler's theorem for homogeneous functions: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -2 u(x,y)$$ - Compute second derivatives and apply the operator: $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = (-2)(-2 -1) u = 6 u(x,y)$$ --- 3. **Function (ii):** $$u(x,y) = \sin^{-1}\left( \frac{(x^4)^{1/5} + (y^4)^{1/5}}{(x^5)^{1/5} + (y^5)^{1/5}} \right) = \sin^{-1}\left( \frac{x^{4/5} + y^{4/5}}{x + y} \right)$$ - Replace $(x,y)$ by $(tx, ty)$: $$u(tx, ty) = \sin^{-1}\left( \frac{t^{4/5} x^{4/5} + t^{4/5} y^{4/5}}{t x + t y} \right) = \sin^{-1}\left( t^{4/5 - 1} \frac{x^{4/5} + y^{4/5}}{x + y} \right) = \sin^{-1}\left( t^{-1/5} \cdot \text{expression} \right)$$ - Since $\sin^{-1}$ is not linear and argument scales by $t^{-1/5}$, $u$ is not homogeneous. - So no degree. - Derivatives are complicated; no simple Euler relation applies. --- 4. **Function (iii):** $$u(x,y) = \frac{x^2 + y^2 + 1}{x + y}$$ - Replace $(x,y)$ by $(tx, ty)$: $$u(tx, ty) = \frac{t^2 (x^2 + y^2) + 1}{t(x + y)} = \frac{t^2 (x^2 + y^2) + 1}{t(x + y)}$$ - This is not equal to $t^k u(x,y)$ for any $k$ because of the $+1$ in numerator. - So $u$ is not homogeneous. --- 5. **Function (iv):** $$u(x,y) = \log \left( \frac{x^2 + y^2}{x + y} \right)$$ - Replace $(x,y)$ by $(tx, ty)$: $$u(tx, ty) = \log \left( \frac{t^2 (x^2 + y^2)}{t (x + y)} \right) = \log \left( t \frac{x^2 + y^2}{x + y} \right) = \log t + \log \left( \frac{x^2 + y^2}{x + y} \right)$$ - Since $u(tx, ty) \neq t^k u(x,y)$, $u$ is not homogeneous. --- **Summary:** - (i) Homogeneous of degree $-2$. - (ii), (iii), (iv) Not homogeneous. **For (i):** $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -2 u(x,y)$$ $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = 6 u(x,y)$$