Question

For a twice continuously differentiable function 𝑔: ℝ → ℝ, define
$u_g(x,y)=\frac{1}{y}\int^y_{-y}g(x+t)dt\ \ \ \text{for}(x,y)\in \R^2, \ \ \ y \gt0.$
Which one of the following holds for all such 𝑔 ?

• A. $\frac{∂^2u_g}{∂x^2}=\frac{2}{y}\frac{∂u_g}{∂y}+\frac{∂^2u_g}{∂y^2}$
• B. $\frac{∂^2u_g}{∂x^2}=\frac{1}{y}\frac{∂u_g}{∂y}+\frac{∂^2u_g}{∂y^2}$
• C. $\frac{∂^2u_g}{∂x^2}=\frac{2}{y}\frac{∂u_g}{∂y}-\frac{∂^2u_g}{∂y^2}$
• D. $\frac{∂^2u_g}{∂x^2}=\frac{1}{y}\frac{∂u_g}{∂y}-\frac{∂^2u_g}{∂y^2}$

Answer 1

Answer: (A)

$\frac{∂^2u_g}{∂x^2}=\frac{2}{y}\frac{∂u_g}{∂y}+\frac{∂^2u_g}{∂y^2}$

Answer 2

The correct option is (A) : $\frac{∂^2u_g}{∂x^2}=\frac{2}{y}\frac{∂u_g}{∂y}+\frac{∂^2u_g}{∂y^2}$.