User:Egm6321.f11.team4.allen/HW3

=''' Problem R3.1 - Show Eq. (1) p.13-2 is exact only if $$k_1=d_1$$ (constant) and Eq. (1) p.12-4 is a particular equation '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-3]]

From [[media:Pea1.f11.mtg11.djvu|Mtg 11-2]]

From [[media:Pea1.f11.mtg12.djvu|Mtg 12-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$. Eq. (2) p.11-2 is $$\displaystyle n(x):=(-1/N)(N_x-M_y)$$. Eq. (1) p.12-4 is $$\displaystyle M+Ny'=[a(x)y+k_x(x)]+\bar b(x)y'=0$$.

Find
$$\displaystyle a)$$Show that the N1-ODE (1) p.13-2 satisfies the condition (2) p.11-2 that an integrating factor $$\displaystyle h(x)$$ can be found to render it exact, only if $$\displaystyle k_1(y)=d_1$$ (constant). $$\displaystyle b)$$Show that (1) p.13-2 includes (1) p.12-4 as a particular case.

Solution
Solved independently. $$\displaystyle a)$$Eq. (1) p. 13-2 can be written as $$\displaystyle N(x,y)y'+M(x,y)=0$$ which yields the following: $$\displaystyle N(x,y)=\bar b(x,y)c(y)$$ $$\displaystyle M(x,y)=a(x)\bar c(x,y)$$. These two equations yield the following when differentiated: $$\displaystyle N_x=\bar b(x,y)c_x(y)+\bar b_x(x,y)c(y)$$ $$\displaystyle M_y=a(x)\bar c_y(x,y)+a_y(x)\bar c(x,y)$$. Now we plug these equations into Eq. (2) p. 11-2 to obtain $$\displaystyle n(x)=\frac{-1}{\bar b(x,y)c(y)}*[\bar b_x(x,y)c(y)-a(x)\bar c_y(x,y)]$$ This simplifies to $$\displaystyle n(x)=\frac{a(x)\bar c_y(x,y)}{\bar b(x,y)c(y)}-\frac{\bar b_x(x,y)c(y)}{\bar b(x,y)c(y)}$$. Since $$\displaystyle \bar b_x(x,y)=b(x,y)$$ & $$\displaystyle \bar c_y(x,y)=c(x,y)$$ cancel out, the formula reduces to $$\displaystyle n(x)=\frac{a(x)}{\bar b(x,y)}-1$$. Rearranging the equation produces $$\displaystyle \bar b(x,y)=\frac{a(x)}{n(x)+1}$$. If you differentiate the above equation by $$\displaystyle y $$, you obtain $$\displaystyle \bar b_y(x,y)=0$$ because everything is a function of $$\displaystyle x $$ or a constant. This shows that $$\displaystyle k_1(y)=d_1$$ (constant) and not a function of $$\displaystyle y$$. $$\displaystyle b)$$In order to show Eq. (1) p. 12-4 is a particular case of Eq. (1) p. 13-2, we must make two assumptions. Those assumptions include $$\displaystyle \bar b(x,y)=\bar b(x)$$ and $$\displaystyle c(y)=C$$(constant). Working first with $$\displaystyle M(x,y) $$, we have $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ (left side is the general case and the right side is the particluar case). With $$\displaystyle c(y)=C $$, Eq. (1) p. 13-3 reduces to $$\displaystyle \bar c(x,y)=Cy+k_2(x)$$. Plug the above equation for $$\displaystyle \bar c(x,y) $$ into the $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ which yields $$\displaystyle a(x)[Cy+k_2(x)]=a(x)y+k_2(x)$$. The left side simplies becase $$\displaystyle C $$ is absorbed into $$\displaystyle a(x) $$ and $$\displaystyle a(x) $$ is absorbed into $$\displaystyle k_2(x) $$. Working last with $$\displaystyle N(x,y) $$, we have $$\displaystyle \bar b(x,y)c(y)=\bar b(x)$$ (left side is the general case and the right side is the particluar case). Since $$\displaystyle c(y)=C $$, it can be absorbed into $$\displaystyle \bar b(x,y)$$. Also, $$\displaystyle \bar b(x,y)=\bar b(x)$$ because it is not a function of $$\displaystyle y $$. This ultimately yields the particluar case, $$\displaystyle \bar b(x)$$.

=''' Problem R*3.2 - Show Eq. (1) p.13-4 is exact or can be made exact by the IFM. Find the integrating factor h '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = 5x^3+2 $$ $$\displaystyle b(x) = x^2 $$ $$\displaystyle c(x) = y^4 $$ $$\displaystyle \bar b(x) = \frac{1}{3}x^3+d_1 $$ $$\displaystyle \bar c(x) = \frac{1}{5}y^5+\sin(x)+d_2 $$ So, after combining the above equations into Eq. (1) p.13-4 we obtain Eq. (1) p.13-4 which is $$\displaystyle (\frac{1}{3}x^3+d_1)y^4y'+(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)=0 $$.

Find
Show that Eq. (1) p. 13-4 either is exact or can be made exact by the IFM. Find the integrating factor $$\displaystyle h $$.

Solution
Solved independently. In order for Eq. (1) p. 13-4 to be exact, it must satisfy both conditions of exactness. The first exactness condition requires the formula to be in the following form: $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (1) p. 13-4 already satisfies the above requirement so it meets this condition. The second exactness condition requires the formula to satisfy the following condition: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)=(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)$$ and $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, to obtain $$\displaystyle M_y(x,y)=5*(x^3+\frac{2}{5})y^4$$ and $$\displaystyle N_x(x,y)=x^2y^4$$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, $$\displaystyle N_x(x,y)=x^2y^4 $$, and $$\displaystyle M_y(x,y)=(5x^3+2)y^4 $$ we have $$\displaystyle h(x)=\exp \int^x -[\frac{5s^3+s^2+2}{\frac{1}{3}s^3+d_1}]ds $$.

= Problem R*3.3 - Find an N1-ODE of the form (1) p. 13-2 and the First Integral, Phi =

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = \sin(x^3) $$. $$\displaystyle b(x) = \cos(x) $$. $$\displaystyle c(x) = \exp(2y) $$.

Find
$$\displaystyle 1) $$ Find an N1-ODE of the form (1) p. 13-2 that is either exact or can be made exact by the IFM. $$\displaystyle 2) $$ Find the first integral $$ \phi (x,y)=k $$.

Solution
Solved independently. $$\displaystyle 1) $$ First solve for $$\displaystyle \bar b(x,y) $$ and $$\displaystyle \bar c(x,y) $$. $$\displaystyle \bar b(x,y) = \int^x \cos(s) ds $$ $$\displaystyle \bar b(x,y) = \sin(x) $$ $$\displaystyle \bar c(x,y) = \int^x \exp(2y) ds $$ $$\displaystyle \bar c(x,y) = \frac{1}{2}*\exp(2y) $$ Fill Eq. (1) p. 13-2 with the known values and you will obtain $$\displaystyle \sin(x)*\exp(2y)*y'+\frac{1}{2}*\sin(x^3)*\exp(2y)=0 $$ This equation simplifies to $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$. So, the above equation satisfies the first condition of exactness, $$\displaystyle N(x,y)y'+M(x,y)=0 $$. However, it fails to satisfy the second condition of exactness, as shown below. The second exactness condition requires the formula to satisfy the following form: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)= \frac{1}{2}*\sin(x^3)*\exp(2y) $$ and $$\displaystyle N(x,y)= \sin(x)*\exp(2y) $$, to obtain $$\displaystyle M_y(x,y)= \sin(x^3)*\exp(2y) $$ and $$\displaystyle N_x(x,y)= \cos(x)*\exp(2y) $$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=\sin(x) $$, $$\displaystyle N_x(x,y)=\cos(x) $$, and $$\displaystyle M_y(x,y)=0 $$ we have $$\displaystyle h(x)=\exp[\int^x [\frac{-1}{\sin(s)}(\cos(s)-0)]ds] $$. This simplies to $$\displaystyle h(x)= \frac{1}{\sin(s)} $$. Multiply $$\displaystyle h(x)= \frac{1}{\sin(s)} $$ and $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$ which yields $$\displaystyle y'+\frac{1}{2}\frac{\sin(x^3)}{\sin(x)}=0 $$. $$\displaystyle 2) $$ Find $$ \phi (x,y)=k $$. $$ \phi_x(x,y)=M(x,y) $$ $$ \phi_y(x,y)=N(x,y) $$ Using the above equations, you can integrate with respect to $$\displaystyle x $$ or $$\displaystyle y $$ which yields $$\displaystyle \phi(x,y)=\frac{-1}{2x^2}\cos(x^3)+k_3 $$ $$\displaystyle \phi(x,y)=\sin(x)y+k_4 $$. Equate the equations above and solve for $$\displaystyle k=k_3+k_4 $$ which is $$\displaystyle \phi(x,y)=\sin(x)y+\frac{1}{2x^2}\cos(x^3)=k $$