Question

if Uxx + Uyy=0 in x^2 + y^2=1 and U = 3+x+y find U(1/2, 1/2)
A)0
B)2
C)3
D)4

Answer 1

ANSWER on Question #79245 – Math – Differential Equations

QUESTION

If

$$U_{xx} + U_{yy} = 0 \quad \text{in} \quad x^2 + y^2 = 1, \quad \text{and} \quad U(x,y) = 3 + x + y$$

Find

$$U\left(\frac{1}{2}, \frac{1}{2}\right)$$

Variants of answer:

A) $U\left(\frac{1}{2}, \frac{1}{2}\right) = 0$

B) $U\left(\frac{1}{2}, \frac{1}{2}\right) = 2$

C) $U\left(\frac{1}{2}, \frac{1}{2}\right) = 3$

D) $U\left(\frac{1}{2}, \frac{1}{2}\right) = 4$

SOLUTION

Hint: The question looks strange, since the explicit form of the function $U(x,y)$ is specified. That is, the first condition: a partial differential equation is not used in any way to solve this problem.

In our case,

$$U(x,y) = 3 + x + y \rightarrow U\left(\frac{1}{2}, \frac{1}{2}\right) = 3 + \frac{1}{2} + \frac{1}{2} = 3 + 0.5 + 0.5 = 4$$

Conclusion,

$$\boxed{U(x,y) = 3 + x + y \rightarrow U\left(\frac{1}{2}, \frac{1}{2}\right) = 4}$$

ANSWER: D) $U\left(\frac{1}{2}, \frac{1}{2}\right) = 4$

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