User:Egm6321.f12.team7.parikh.a/HW18.1

=R*3.14 - Find solution for h(x,y) =

Given
$$\displaystyle g=\cancel{xp^2}+\cancel{yp} = \phi_x + \phi_y p = \underbrace{(h_x + \cancel{yp})}_{\phi_x}+\underbrace{(h_y+\cancel{xp})}_{\phi_y}p $$ $$\displaystyle p(x)= y'(x) $$

Find
$$\displaystyle h_x + h_y p = 0$$3.14.1

Find h(x,y).

Solution
$$\displaystyle h_x + h_y p = 0 $$ where

$$\displaystyle h_x = \frac {\partial h}{\partial x}$$ $$\displaystyle h_y = \frac {\partial h}{\partial y}$$ $$\displaystyle p = \frac {dy}{dx} $$ The eq 3.14.1 is following that

$$ \displaystyle \frac {\partial h(x,y)}{\partial x}+\frac {\partial h(x,y)}{\partial y} \frac {dy}{dx} = \frac {d}{dx} h(x,y) = 0$$ Since $$\displaystyle \frac {d}{dx} h(x,y)$$ is 0, h(x,y) is only the function of y.

Rearranging and integrating both sides:

$$ dh(x,y) = 0dx $$

$$ \int dh(x,y) = \int 0dx $$

$$ h(x,y) = C $$

Therefore, h has to be a constant.