User:Egm6321.f09.team3.mooney/HW2

Problem #5
Show that the second exactness condition is satisfied for $$x*y*y' + x*(y')^2 + y*y' =0$$

The second exactness condition is described by the following 2 equations

Equation 1

 * $${f}_{xx} + 2*p*{f}_{xy} + p^2 * {f}_{yy} = {g}_{xp} + p*{g}_{yp}-{g}_{y}$$

Equation 2

 * $${f}_{xp} + p*{f}_{yp} + 2*{f}_{y} = {g}_{pp}$$

solving for the individual components by taking the appropriate partial derivatives gives the following values:
 * $${f}_{xx} = 0$$
 * $${f}_{xy} = 1 $$
 * $${f}_{yy} = 0 $$
 * $${f}_{xp} = 0 $$
 * $${f}_{yp} = 0 $$
 * $${f}_{y} = x  $$
 * $${g}_{xp} = 2p $$
 * $${g}_{yp} = 1 $$
 * $${g}_{pp} = 2x $$
 * $${g}_{y} = p  $$

Therefore to check for exactness we can substitute the calculated values: Equation 1 becomes:
 * $$0 + 2*p*(1) + p^2*(0) = 2*p + p*(1) -p$$ ==> $$2*p = 2*p$$ this satisfies the first part of the condition

Equation 2 becomes:
 * $$0 + p*(0) + 2*x = 2*x$$ ==> $$2*x = 2*x$$ satisfying the second part of the condition

Therefore the equation $$x*y*y' + x*(y')^2 + y*y' =0$$ is exact.

Problem #6
Given $$f = {\phi}_{p}$$ and $$g = {\phi}_{x} + {\phi}_{y}*p$$, derive the second condition of exact through differentiation.

Knowing that partial derivates taken in order provide equal results ($$ {f}_{xy} = {f}_{yx}$$), then we can right the following:


 * $$g = {\phi}_{x} + {\phi}_{y}*p$$


 * $$\frac{\partial g}{\partial p} = {{\phi}_{p}}_{x} + {\phi}_{y} + {\phi}_{yp}*p = {f}_{x} + {\phi}_{y} + {f}_{y}*p$$


 * $$\frac{\partial^2 g}{\partial p^2} = {f}_{xp} + {f}_{y} + {f}_{y} + p*{f}_{yp}$$


 * $${g}_{pp} = {f}_{xp} + p*{f}_{yp} + 2*{f}_{y}$$

The form matches the second criteria of the second exactness illustrated agreement with proposed solution. QED!

Problem #7
Using $${\phi}_{xy} = {\phi}_{yx}$$, $$f = {\phi}_{p}$$ , and $$g = {\phi}_{x} + {\phi}_{yp}$$ derive $${f}_{xx} + 2*p*{f}_{xy} + p^2 * {f}_{yy} = {g}_{xp} + p*{g}_{yp}-{g}_{y}$$