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R*4.4 Question from meeting 21, pages 5-6: Given (2)-(3) from p. 21-4 $$G = \underbrace{\underbrace{R(x)}_{\displaystyle a_0(x)} y}_{\displaystyle \ne \phi_x} + \underbrace{\underbrace{Q(x)}_{\displaystyle a_1(x)} y'}_{\displaystyle \ne \phi_y} + \underbrace{\underbrace{P(x)}_{\displaystyle a_2(x)} y''}_{\displaystyle = \phi_p}$$ $$\displaystyle (Equation\;4.4.1)$$ $$\phi_x(x,y,p) + \phi_y(x,y,p) \,y' = R(x)y + Q(x)y'$$ $$\displaystyle (Equation\;4.4.2)$$  Find that the solution gives (1) from p. 21-5 $$\phi(x,y,p) = P(x)p + T(x)y + k$$  $$\displaystyle (Equation\;4.4.3)$$

From lecture notes [[media:pea1.f11.mtg21.djvu|Mtg 21]]

Solution
The equations (2) and (3) are within the category of L2-ODE-VC. The key is to use the substitution $$p = y'$$ for the derivations and $$p' = y''$$.The first step is to prove the exactness of the equations below. There are two exactness conditions for the N2-ODE. Note that the functions R and Q are of x but not y. This affects the 2nd exactness condition and subsequent derivations and integrations.

$$G = \underbrace{\underbrace{R(x)}_{\displaystyle a_0(x)} y}_{\displaystyle \ne \phi_x} + \underbrace{\underbrace{Q(x)}_{\displaystyle a_1(x)} y'}_{\displaystyle \ne \phi_y} + \underbrace{\underbrace{P(x)}_{\displaystyle a_2(x)} y''}_{\displaystyle = \phi_p}$$ $$\displaystyle (Equation\;4.4.1)$$

$$g(x,y,p) = \phi_x(x,y,p) + \phi_y(x,y,p) \,y' = R(x)y + Q(x)y'$$ $$\displaystyle (Equation\;4.4.2$$  The function with p=y' can be rearranged.  $$g(x,y,p) = \phi_x(x,p) + \phi_y(x,p) \,y' = R(x)y + Q(x)p$$ $$\displaystyle (Equation\;4.4.3)$$

The equations satisfy the following form of the 1st exactness condition. $$G = \frac{d \phi}{dx} = \phi_x + \phi_y \,y'+ \phi_p \,y'' $$ $$\displaystyle (Equation\;4.4.4)$$ The 2nd exactness condition is derived below. First, assume that $$f(x,p) := \phi_p(x,p)= P(x)$$ Then the 2nd exactness condition can be applied satisfactorily. $$f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y$$ $$f_{xp} + pf_{yp} + 2f_y = g_{pp}$$ $$\displaystyle (Equation\;4.4.5)$$ This can be shown by placing the values of R and Q into the equation for 2nd exactness. $$P_{xx} + 0 + 0 = g_{xp} + pg_{yp} - g_y = Q_{x} - R$$ The next step is to find the integration factor h(x) from the equation below. $$g(x,yp) := \phi_x + \phi_y \,y' = \phi_x(x,p) + \phi_y(x,p) \,p = R(x)y + Q(x)p$$ Recall that the integration factor in this case is a function of x, so the following equation from meeting 11 can be used. $$\frac{h_x}{h} = -\frac{1}{N}(N_x - M_y) =: \color{blue}{n(x)}$$ The integration is simple. <p style="font-size:125%" align="center">$$h(x) = \int \frac{h_x}{h} = -\int \frac{1}{N}(N_x - M_y) = -\int\frac{1}{Q(x)}(Q_x - R)= T(x)y + k$$ Subsequently, the final solution is equated below. <p style="font-size:125%" align="center">$$\phi(x,y,p) = h(x) + \int(f,x,y,p) \,dp = P(x)p + T(x)y + k$$ References for R4.4 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 16) [[media:pea1.f11.mtg16.djvu|Mtg 16]] 22 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 21) [[media:pea1.f11.mtg21.djvu|Mtg 21]] 04 Oct 2011.