User:Egm6321.f12.team4.hong/Report2

Problem R* 2.6: Proof of Mixed Derivative Theorem (Clairaut's Theorem)
I solved this problem on my own.

Statement
Review calculus, and find the minimum degree of differentiability of the function such that $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$ is satisfied. State the full theorem and provide a proof.

Solution
Theorem $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$ when, $$ phi(x,y), phi_{x}, phi_{y}, phi_{xy}, phi_{yx} $$ are defined throughout an open and are all continuous throughout an open region.

Proof Equality of this theorem can be established by mean value theorem. Let h, k be arbitrary real numbers,

Let's define a function and apply the mean value theorem,

From the above equations,

This time let there be a function similar to the second equation but instead of y being the variable, let x be the variable.

Degree of Differentiability

''Looking at the theorem and the proof above, in order to satisfy the equality, the function must be differentiated at least three times. Therefore the minimum Degree of Differentiability for Mixed Derivative Theorem is three.''

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reference -- Thomas's Calculus, International Edition (George B. Thomas, Jr): pg. AP-23 ~ AP-25

Problem R* 2.9: Finding Integrating Factor h(x,y) when h_{x}(x,y)=0
I solved this problem on my own.

Statement
Suppose $$ h_{x}(x,y)=0,$$thus h is a function of y only; then $$ h_{x}N - h_{y}M + h(N_{x} - M_{y})=0 $$ becomes $$ \frac{hy}{h}=\frac{1}{M}(N_{x}-M_{y})=:m(y) $$ Find h using the above equation.

Solution
From the class note 11-2 it is written that,

If $$ h_{y}(x,y) = 0 $$ then,

In the same manner, this problem can be solved.

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Author and References

 * Solved and Typed by -- Seong Hyeon Hong